Problem 7.8. In a chess tournament, 30 chess players participated, each playing against each other exactly once. A win was awarded 1 point, a draw - $1 / 2$, and a loss - 0. What is the maximum number of chess players who could have ended up with exactly 5 points by the end of the tournament?
Answer: 11.

Solution. Let $N-$ be the number of chess players with 5 points.

Assume $N \geqslant 12$. In each game between two playing chess players, 1 point is played, so the sum of points of all chess players who scored 5 points at the end of the tournament is no less than $\frac{N \cdot(N-1)}{2}$. Then, by the pigeonhole principle, there will be one among them with points no less than $\frac{\stackrel{2}{N-1}}{2} \geqslant 5.5$, a contradiction. Therefore, $N \leqslant 11$.

We will show that the value $N=11$ is possible. Divide the 30 chess players into two groups: the first with 11 people, and the second with 19. Let each person in each group play to a draw with each other, and also let each person in the first group lose to each person in the second group. It is easy to see that exactly 5 points were scored by all chess players in the first group, and there are 11 of them.