4. In a school chess tournament, boys and girls competed, with the number of boys being five times the number of girls. According to the tournament rules, each chess player played against every other player twice. How many players in total participated if it is known that the boys scored exactly twice as many points in total as the girls? (1 point is awarded for a win, -0.5 points for a draw, and 0 points for a loss.)
Answer: 6 players.

Solution. Let $d$ girls and $5d$ boys participate in the tournament. Then the total number of players was $d + 5d = 6d$; playing two matches each with every other, they played a total of $2 \cdot \frac{1}{2} \cdot 6d(6d-1) = 6d(6d-1)$ matches. Since each match awards one point, the total number of points scored by all participants is also $6d(6d-1)$. Of these, two-thirds of the points belong to the boys, and one-third to the girls, meaning the girls scored $\frac{1}{3} \cdot 6d(6d-1) = 2d(6d-1)$ points.

Note that if each girl won against all the boys, the girls together could have scored a maximum of $2 \cdot d \cdot 5d = 10d^2$ points, and playing against each other, the girls distributed $2 \cdot \frac{1}{2} \cdot d(d-1)$ points. Therefore, the maximum number of points the girls could have scored is $10d^2 + d(d-1) = 11d^2 - d$. Thus,

$$
2d(6d-1) \leq 11d^2 - d \quad \Longleftrightarrow \quad d^2 \leq d
$$

Therefore, there could not have been more than one girl. If there was one girl, then there were five boys, making a total of 6 players. Six players played a total of $6 \cdot 5 = 30$ matches and awarded 30 points. The girl scored 10 points, winning two matches against each of the five boys. Playing against each other, the boys distributed the remaining 20 points.

Criteria. Only the answer - 0 points. Correct example - 2 points. Calculation of the total number of points scored by all girls - 2 points. Estimation of the number of points scored by the girls - 4 points. Criteria are not cumulative. Full solution - 7 points.