Why Chinese Komi should increase by 2 point steps
Robert put advanced explanations into brackets. I moved them to the bottom as footnotes --Arno
Ted. S. asked: I still don't understand why Chinese rules can't have a 6.5 full count komi (3.25 zi) which isn't substantially different from a 5.5 (2.75 zi). Why does Chinese komi have to increase in two-point intervals?
The Chinese use half count komi but since you are more familiar with full count komi, I shall rely figures on it. Simply speaking, your question is about the difference between 5.5 and 6.5 komi.
The Chinese score stones plus empty intersections.
Let us first understand how the Chinese score:
5 black intersections score for Black. 10 empty intersections score for Black. 5 white intersections score for White. 5 empty intersections score for White. 0 komi score for White.
The sum (positive values for Black) is: 5 + 10 - 5 - 5 - 0 = 5. Hence Black wins by 5 points. (Using full point counting for your convenience.)
This article is about parity - even parity or odd parity. We need to think about parity because it is a parity argument that distinguishes the characteristics of a 5.5 komi from a 6.5 komi.
What is the effect of komi on diagram 1?
First we need to understand the parity of the board in diagram 1. It has 25 intersections. This is an odd number. The parity of the board is odd. This is the same parity as the parity of a 19x19 board. Therefore a parity argument illustrated on a 5x5 board will also be valid on a 19x19 board.
Compared to diagram 1, in diagram 2 white gets 5.5 points more. Therefore, while in diagram 1 Black wins by 5 points, in diagram 2 White wins by 0.5 points. In particular we note that White wins.
Now White wins by 1.5 points. In particular we note that White wins. Despite a komi change from 5.5 to 6.5 the winner does not change!
This is a general rule for a board of an odd size! 
The truth is, of course, there is no general rule without exception. There are two exceptions:
- handicap stones
We disregard handicaps because in Chinese professional play there is (usually) no handicap; all games are even games. (This "even" in the term "even game" does not express a parity...)
What we need to discuss is sekis because sekis do occur in professional games. In terms of parity, there are two types of seki: even seki and odd seki.
There is a seki on the board. It has two neutral intersections that do not score. This is an even number of neutral intersections. We have an even seki.
9 black intersections score for Black. 5 empty intersections score for Black. 9 white intersections score for White. 0 empty intersections score for White. 5.5 komi score for White.
Sum: 9 + 5 - 9 - 0 - 5.5 = -0.5. White wins by 0.5 points.
Compared to diagram 4, white has 1 point more and thus wins by 1.5. Independently of whether the komi is 5.5 or 6.5, the winner is the same!
Even sekis with an even number of neutral intersections do not change the winner if the komi is increased from 5.5 to 6.5.
Now we have an odd seki with an odd number of neutral intersections, namely 1 neutral intersection. 1 is an odd number.
9 black intersections score for Black. 6 empty intersections score for Black. 8 white intersections score for White. 1 empty intersections score for White. 5.5 komi score for White.
Sum: 9 + 6 - 8 - 1 - 5.5 = 0.5. Black wins by 0.5. Black is the winner.
Compared to diagram 6, white has 1 point more. White wins by 0.5. White is the winner.
Increasing the komi from 5.5 to 6.5 has changed the winner!
It has required one odd seki to achieve this! Such odd sekis are scarce in practice.
Therefore the Chinese would not increase komi from 5.5 to 6.5 but rather from 5.5 to 7.5! A komi increment from 5.5 to 6.5 would be meaningless in most of the games.
(edited by ArnoHollosi)
 Instead of these two values we could also study the same for other values like 1.5 / 2.5 or 3.5 / 4.5 or 7.5 / 8.5. We study it for 5.5 / 6.5 in particular because the previous komi in China has been 5.5.
 A player's own stones and empty intersections surrounded only by his own stones, of course. This type of scoring is called area scoring. I frequently use that term so if you want to understand what I say in other articles, first you should understand this term. It is much more convenient for a frequent writer to use a short term like "area scoring" than to declare each time that a discussed scoring is defined as own stones plus empty intersections surrounded only by own stones, that this scoring is used in China, by Taiwanese professionals, in USA tournaments, in New Zealand, and in every other tournament using Chinese, Ing, American, New Zealand, or similar rules, that that way of scoring is sometimes informally called Chinese scoring, Chinese counting, the Chinese way of counting, the Chinese counting method, and that Japanese or Korean rules use a different scoring method. The purpose of a term is to make a text readable; hardly any of my texts about scoring rules would be readable if I used the previous sentence instead of the short term "area scoring".
Also, I could not type my average of 1,000 (?) rec.games.go articles per year if I did not use any such specialized terms. When you say that you never seem to understand my explanations, it might also depend on your eagerness to understand the terms that I use. E.g., have you tried to find out what "standard area komi" means? This is what the discussion is all about. E.g., 5.5 is a standard area komi while 6.5 is not a standard area komi. I shall explain the reason again below.
 If you want to understand it in general, then you are required to use and understand some mathematics. Am afraid that I shall not use mathematics because then you might complain not to understand me again. So all I can do is to suggest that you try for yourself many example positions on a 5x5 board and see whether you could break this general rule. You cannot.
 It does not matter how often white might have passed during the game or how many stones he might have lost during it. This does not change the fact that the players played in alternation.
Whenever white passes and a black stone instead of a white stone finds its way onto the board, we have one white stone fewer and one black stone more; this is a difference of 2 points; this is an even difference; with such the parity does not change. Whenever a white stone is removed from the board and the surrounded intersection it is removed from scores for black, we have one white stone fewer and one emtpy intersection scoring for black more; this is a difference of 2 points; this is an even difference; with such the parity does not change.
However, a handicap can change the parity, although a handicap with H handicap stones can be understood as H - 1 black plays on the board in alternation with H - 1 white passes before the Hth black stone on the board starts regular alternation. Why? It is not the passes themselves what changes the parity of a score on the board at the game end but it is the parity of the length of alternation of moves throughout a game. During one game the board parity does not change; it is constantly odd. Simplifying, we assume that the even seki parity also does not change.
Hence in case of an even sum of empty black plus empty white intersections on the board black always plays the last stone on some neutral intersection, if there are any immediately before the final passes, and in case of an odd sum of empty black plus empty white intersections on the board white always plays the last stone on some neutral intersection, if any.
However, every handicap stone changes the number of remaining intersections on the board by 1 and therefore also changes the parity. Since every 2 handicap stones the parity is the same again, it is sufficient to consider the first handicap stone of all handicap stones (i.e. the handicap stone before any pass occurs in alternation with further handicap plays). The first handicap stone is followed by either an odd or an even number of further handicap stones before the last handicap stone starts regular alternation.
E.g., if the handicap is 2, then we have 0 further handicap stones after the first handicap stone and before the last handicap stone. The parity of the further handicap stones is even, the parity of (the number of played stones of) the first handicap stone is odd, the parity of the sum of all handicap stones before the last one, which starts regular alternation, is odd. Since it is odd, this changes the parity of the length of remaining moves in the game and therefore in a game with 2 handicap stones white plays the last neutral point in the game that ends with an even sum of empty black plus empty white intersections on the board. (If white tenukis to fill his territory, then there the parity changes and as a natural consequence black would be the happy player to get the last "neutral" point.)
E.g., if the handicap is 3, then we have 1 further handicap stones after the first handicap stone and before the last handicap stone. The parity of the further handicap stones is odd, the parity of the first handicap stone is odd, and the parity of the sum of all handicap stones before the last handicap stone is even. Since it is even, a handicap 3 does not change black as the last player to occupy a neutral point in a game that ends with an even sum of empty black plus empty white intersections on the board, quite as in a no handicap game.
Suppose in diagram 4 with its odd sum of scoring empty intersections we would have a 4 stones handicap game. At the end there are 9 black stones and 9 white stones on the board. Since the game would have started with
black play white pass black play white pass black play white pass black play
, it must have ended with
black pass white play black pass white play black pass white play black pass white pass
It is white who plays the last neutral point. Contrarily we expected black to play the last neutral point because the parity of the scoring empty intersections is odd, the handicap parity effect is odd. Odd and odd together should cancel out and we should get the same behaviour as in an even game. What has happened?
The game end has been strange. White was allowed an extra play at the game end that changed the parity on the board again. Normally, black should not pass too early at the end at be the player to occupy the last neutral point:
Diagram 4b with 4 stones handicap of which move 1 is the 4th handicap stone; then alternation continues; the players take points rather than passing too early during the endgame
This results in: