I think I'm misunderstanding, because the problem seems wrong to me. Point (a) is ok: Let's assume each player has a different number of ties. All players must have a number of ties with the same parity, because the parity of the number of ties tells us if the total score, which is common to all players, has form $n$ or $\frac {2n+1}2$. The number of ties of a player ranges from $0$ to $2n-2$. We must have $n$ different numbers from this set, all with the same parity, so they must be $0,\ 2,\ ..,\ 2n-2$. This means that there is a player $A$ who has tied in all his matches, and there is a player $B$ with no ties. What happens when $A$ plays with $B$? This obviously yields a contradiction, so there must be two players with the same number of ties. As for point (b), I think it's very easy to give a counterexample: Let's assume we only have two players in the tournament, $A,\ B$. When $A$ plays with the white pieces he wins, and when he plays with the black pieces he loses. $A$ has $0$ lost games when playing with the white pieces, while $B$ has $1$ lost game when playing with the black pieces. However, the total score is $1$ for both $A$ and $B$, so the conditions of the problem are satisfied. This sort of counterexample can easily be extended to more players, so a condition like "at least three players" doesn't work. What am I doing wrong?

you made the same mistake as most contestants did. probably you are not very fond of chess anyway, the thing is that the problem ask to prove that there exist two players which have the same number of lost games when playing with the white pieces. in fact in your example, both A and B have 0 losses when playing with the white pieces.

Oh my! That's a very stupid mistake Sorry..

where´s the slution of item b? please,give me a solution

Chavez, be patient! Or, better, go solve the question yourself

Suppose that there are $n$ players. Note that there are $n(n-1)$ games in total, hence every player got $n-1$ points. a) Since this is an integer, we see that every player had an even number of tied games. Since everyone played $2(n-1)$ games, there would be exactly one player with $2k$ draws for $k=0,1\ldots, n-1$ if the statement was wrong. We obtain a contradiction by considering the games between the player with $0$ draws and the one with $2(n-1)$ draws. b) Pretty similar. If the statement was wrong, there would be exactly one player with $k$ lost white games for $k=0,\ldots, n-1.$ Consider the player with $0$ losses as white, who must have lost all the remaining games (as he otherwise would end up with more than $n-1$ points). But then he also lost his game with black against the player with $n-1$ losses at white, i.e with $0$ wins, impossible.