1. In a volleyball tournament, $n$ teams from city $A$ and $2 n$ teams from city $B$ participated. Each team played exactly one game against each other team. The ratio of the number of wins by teams from city $B$ to the number of wins by teams from city $A$ is $3: 4$. Find $n$, given that there were no draws in the tournament.
(12 points) Solution. The number of games in which only teams from city $A$ participated is $\frac{(n-1) n}{2}$. In these games, teams from city $A$ won $\frac{(n-1) n}{2}$ games. The number of games in which only teams from city $B$ participated is $\frac{(2 n-1) 2 n}{2}$. In these games, teams from city $B$ won $(2 n-1) n$ games. The number of matches between teams from city $A$ and teams from city $B$ is $2 n^{2}$. Let $m$ be the number of wins in these matches by teams from city $A$, then teams from city $B$ won $2 n^{2}-m$ games in these matches. In total, teams from city $A$ won $\frac{(n-1) n}{2}+m$ games, and teams from city $B$ won $(2 n-1) n+2 n^{2}-m$ games. According to the condition, we have $\frac{\frac{(n-1) n}{2}+m}{(2 n-1) n+2 n^{2}-m}=\frac{4}{3}$, $3 n^{2}-3 n+6 m=32 n^{2}-8 n-8 m, 14 m=29 n^{2}-5 n, m=\frac{29 n^{2}-5 n}{14}, \frac{29 n^{2}-5 n}{14} \leq 2 n^{2}, n^{2}-5 n \leq 0$. The number $n$ can be $1,2,3,4,5$. By substituting into the equation $m=\frac{29 n^{2}-5 n}{14}$, we get that $m$ is an integer only when $n=5$.

Answer: 5.