, 2022

Proof. We prove by contradiction: assume the conclusion is false. Suppose that the teams $P_1, \cdots, P_k$ all have won less than $\frac{n}{4}$ games, while the remaining $3n-k$ teams, $Q_1, \cdots, Q_{3n-k}$ all have won at least $\frac{n}{4}$ games. Based on the assumption of proof by contradiction, the number of defeats for $Q_1, \cdots, Q_{3n-k}$ is also less than $\frac{n}{4}$. Since the total number of victories for all teams is $3n^2$, there must be at least one team with no fewer than $n$ victories, implying that $3n-k > 0$.
The number of matches between teams $P_1, \cdots, P_k$ cannot exceed the sum of their victories, so it is not more than $\frac{nk}{4}$. In a similar fashion, the number of matches between teams $Q_1, \cdots, Q_{3n-k}$ cannot exceed the sum of their defeats, so it is less than $\frac{n(3n-k)}{4}$ (using the fact that $3n-k > 0$).
The number of matches between the two sets $P_1, \cdots, P_k$ and $Q_1, \cdots, Q_{3n-k}$ is at most $k(3n-k)$ games. Consequently, the total number of matches $N$ satisfies:
$$N < \frac{nk}{4} + \frac{n(3n-k)}{4} + k(3n-k) \le \frac{3n^2}{4} + \frac{9n^2}{4} = 3n^2,$$
which contradicts the given condition. Therefore, the assumption of proof by contradiction is incorrect, and the original proposition is true. $\square$