In a round-robin tournament, 23 teams participated. Each team played exactly once against all the others. We say that 3 teams have cycled victories if, considering only their games against each other, each of them won exactly once. What is the maximum number of cycled victories that could have occurred during the tournament?
Let's solve the problem more generally for $n$ teams, where $n$ is an odd number. We will represent each team with a point. After a match between two teams, an arrow should point from the losing team to the winning team. (In this context, we can assume that there has never been a tie.)

If the teams $A$, $B$, and $C$ have beaten each other in a cycle, which can happen in two ways, then the triangle $ABC$ is called directed. Otherwise, the triangle is not directed.

Provide an estimate for the number of non-directed triangles.

In every non-directed triangle, there is exactly one vertex such that the team represented by this vertex has beaten the other two. We will count the non-directed triangles according to the "winning" vertex.

Let the teams be represented by the points $A_{1}, A_{2}, \ldots, A_{n}$, and let $d_{i}$ denote the number of wins for $A_{i}$. Since there were a total of $\binom{n}{2}$ matches, $\sum_{i=1}^{n} d_{i}=\frac{n(n-1)}{2}$. Then, $\binom{d_{i}}{2}$ non-directed triangles are associated with the vertex $A_{i}$, since we need to choose two from the teams it has beaten.

Thus, the total number of non-directed triangles is: $\sum_{i=1}^{n}\binom{d_{i}}{2}$.

By the inequality between the arithmetic mean and the quadratic mean:

$$
\begin{gathered}
\sum_{i=1}^{n}\binom{d_{i}}{2}=\frac{1}{2} \sum_{i=1}^{n} d_{i}^{2}-\frac{1}{2} \sum_{i=1}^{n} d_{i}=\frac{1}{2} n \frac{\sum_{i=1}^{n} d_{i}^{2}}{n}-\frac{1}{2} \sum_{i=1}^{n} d_{i} \geq \frac{1}{2} n\left(\frac{\sum_{i=1}^{n} d_{i}}{n}\right)^{2}-\frac{1}{2} \sum_{i=1}^{n} d_{i}= \\
=\frac{1}{2} n\left(\frac{n-1}{2}\right)^{2}-\frac{1}{2} \frac{n(n-1)}{2}=\frac{n^{3}-4 n^{2}+3 n}{8}
\end{gathered}
$$

Can equality hold here? Yes, if we can provide a directed complete graph for which

$$
d_{1}=d_{2}=\cdots=d_{n}=\frac{n-1}{2}
$$

Let $n=2 k+1$. Then one possible construction is as follows:

If $1 \leq i<j \leq 2 k+1$ and $j-i$ is greater than $k$, then $A_{j}$ beat $A_{i}$, otherwise $A_{i}$ beat $A_{j}$. It is easy to verify that in this case,

$$
d_{1}=d_{2}=\cdots=d_{n}=\frac{n-1}{2}
$$

Since there are $\binom{n}{3}$ triangles in total, the number of directed triangles in our case is

$$
\binom{n}{3}-\frac{n^{3}-4 n^{2}+3 n}{8}=\frac{n(n-1)(n-2)}{6}-\frac{n^{3}-4 n^{2}+3 n}{8}=\frac{n^{3}-n}{24}
$$

For the case $n=23$, this is 506, so at most 506 cycles of wins could have occurred during the tournament.

Based on the work of András Juhász (Fazekas M. Főv. Gyak. Gimn., 10th grade)