In a table tennis tournament, $n$ Asian and $2 n$ European players participated. The number of matches won by the Europeans is 5/7 of the number of matches won by the Asians. How many players participated in the tournament?
In table tennis, there are no ties, so each match has a winner. The number of matches is distributed as follows. The number of matches where two Asians played against each other is exactly $\frac{n(n-1)}{2}$; in these matches, an Asian always won. The Europeans played $\frac{2 n(2 n-1)}{2}$ matches against each other; in these, a European always won. The number of "mixed" matches is $2 n \cdot n=2 n^{2}$. Suppose that out of these, a European won $x$ times and an Asian won $2 n^{2}-x$ times.

According to the condition,

$$
\frac{5}{7} \cdot\left(\frac{n(n-1)}{2}+2 n^{2}-x\right)=\frac{2 n(2 n-1)}{2}+x
$$

holds; after rearrangement, this yields

$$
n(3-n)=8 x
$$

We know that $n>0$ and $x$ is a non-negative integer. The latter implies $n \leq 3$, so only $n=1,2,3$ are possible. However, only in the case of $n=3$ is $n(3-n)$ divisible by 8. Therefore, $n=3$, meaning nine players participated in the tournament.

Based on the work of András Kobazek (Miskolc, Földes F. Gimn. II. o. t.)