In a chess tournament, $n$ women and $2n$ men participated. Everyone played exactly once with everyone else. There were no draws, and the number of games won by the women is in the ratio of $7:5$ to the number of games won by the men. What is the value of $n$?
Each of the $n$ women played with each of the $n-1$ other women and $2 n$ men. This amounts to a total of $n(n-1)$ games between women and $2 n^{2}$ mixed games. The games between women were counted twice (once for each participant), so the women played only $\frac{1}{2} n(n-1)$ games among themselves.

In these games, of course, the women always won, just as the men always won in the games where both participants were men. The number of the latter is similarly $\frac{1}{2} f(f-1)$, where $f=2 n$ is the number of men. If the women won $N$ of the mixed games and the men won $F$, then on the one hand

$$
N+F=2 n^{2}
$$

and on the other hand

$$
\left[\frac{1}{2} n(n-1)+N\right]:[n(2 n-1)+F]=7: 5
$$

From these two conditions, the usual transformations yield the expressions for the unknowns $N, F$ as

$$
N=\frac{n(17 n-3)}{8}, \quad F=\frac{n(3-n)}{8}
$$

The latter is only non-negative and an integer if $n=3$, at which point the value of $N$ is also appropriate. Therefore, $n=3$.