Let $N \geqslant 3$ be an odd integer. $N$ tennis players take part in a league. Before the league starts, a committee ranks the players in some order based on perceived skill. During the league, each pair of players plays exactly one match, and each match has one winner. A match is considered an upset if the winner had a lower initial ranking than the loser. At the end of the league, the players are ranked according to number of wins, with the initial ranking used to rank players with the same number of wins. It turns out that the final ranking is the same as the initial ranking. What is the largest possible number of upsets?
(United Kingdom)
Answer: $\frac{(N-1)(3 N-1)}{8}$.
Suppose the players are ranked $1,2, \ldots, N=2 n+1$, where 1 is the highest ranking.
For $k \leqslant n$, the player ranked $k$ could have beaten at most $k-1$ players with a higher ranking. Thus the top $n$ players could have made at most $\sum_{k=1}^{n}(k-1)=\frac{n(n-1)}{2}$ upsets. On the other hand, the average score of all $2 n+1$ players is $n$, so the average score of the bottom $n+1$ players is not more than $n$, which implies that these $n+1$ players have at most $n(n+1)$ wins in total. Hence the total number of upsets is at most

$$
\frac{n(n-1)}{2}+n(n+1)=\frac{n(3 n+1)}{2}=\frac{(N-1)(3 N-1)}{8} .
$$

An example can be constructed as follows. Suppose that, for $1 \leqslant i \leqslant 2 n+1$, the player ranked $i$ beats the players ranked $i-1, i-2, \ldots, i-n$ (the rankings are counted modulo $N)$ and loses to the rest of the players. Thus each player has exactly $n$ wins. The player ranked $i$ for $i \leqslant n$ made $i-1$ upsets, whereas the player ranked $i$ for $i>n$ made $n$ upsets, so the total number of upsets is exactly $\sum_{i=1}^{n}(i-1)+(n+1) n=\frac{n(3 n+1)}{2}$.