Twenty-five tennis players are numbered by the numbers $1,2,...,25$. The players are divided into five teams with five players on each team in such a way that the sum of the numbers of the players on each team equals $65$. At a tournament each player plays matches against all other players except those included in his/her own team. After the tournament it turns out that all matches were won by the player with higher number. We say that team $X$ wins against team $Y$ if the players on team $X$ won more matches than they lost against the players on team $Y$. Prove that there exist three teams $A,B,C$, such that $A$ wins against $B$, $B$ wins against $C$, $C$ wins against $A$
Consider the underlying directed graph, where $X \to Y$ if and only if $X$ beats $Y$. The only way that there is no directed triangle is if there is a total ordering, in particular, if there is a team which wins against all other teams.
Now I show that every team must lose at least once. For each element $X$, let $v(x)$ be the number of victories that it scores across all matches. For a team $S$, let $v(S)$ be the sum of $v(x)$ over $x \in S$.
Then we see that for each team $v(S) \leq 65-5-10=50$. (An element $k$ can score at most $(k-1)$ victories, and we can subtract the 10 victories within the elements of $S$). This means that against one of the other 4 teams, $S$ must score at most 12, and hence $S$ must lose at least once.