Problem 11.4. In a sports school, 55 people are training, each of whom is either a tennis player or a chess player. It is known that there are no four chess players who would have the same number of friends among the tennis players. What is the maximum number of chess players that can train in this school?
Answer: 42.

Solution. Let there be $a$ tennis players and $55-a$ chess players in the school. Each of the chess players has a number of tennis player friends that is no less than 0 and no more than $a$, meaning it can take on $a+1$ values. If there were more than $3(a+1)$ chess players, by the pigeonhole principle, there would be four of them with the same number of tennis player friends. Therefore, the number of chess players cannot exceed $3(a+1)$, leading to the inequality $55-a \leqslant 3(a+1)$. Solving this, we get $a \geqslant 13$, so $55-a \leqslant 55-13=42$.

Note also that there could have been exactly 42 chess players: for each integer $0 \leqslant k \leqslant 13$, let some three chess players have exactly $k$ arbitrary tennis player friends.