2. In a chess tournament, everyone played against each other once. The winner won half of the games and drew the other half. It turned out that he scored 13 times fewer points than all the others. (1 point for a win, 0.5 for a draw, 0 for a loss.) How many chess players were there in the tournament?
Answer: 21 chess players.

Solution. If the number of participants is $n$, then each played $n-1$ games. The winner won half of the games and scored $\frac{1}{2}(n-1)$ points. The winner drew half of the games and scored another $\frac{1}{4}(n-1)$ points. In total, the winner scored $\frac{1}{2}(n-1)+\frac{1}{4}(n-1)=\frac{3}{4}(n-1)$ points.

In the tournament, $\frac{1}{2} n(n-1)$ games were played. In each game, the players played for one point, so they scored a total of $\frac{1}{2} n(n-1)$ points. Therefore,

$$
13 \cdot \frac{3(n-1)}{4}=\frac{n(n-1)}{2}-\frac{3(n-1)}{4}
$$

from which $n=21$. Thus, in a tournament of 21 chess players, the winner played 20 games and scored $10+5=15$ points, and the rest scored $\frac{1}{2} \cdot 21(21-1)-15=195$ points, which is 13 times more than the winner.

Criteria. Only the answer - 0 points. Proven that for $n=21$ the tournament meets the condition, but not justified that there are no other solutions - 2 points. The equation for the number of participants is correctly set up - 5 points. Complete solution - 7 points.