Example 6 In a competition with $2 n+1$ teams, each team played a match against every other team, and each match had a winner. If $A$ beats $B$, $B$ beats $C$, and $C$ beats $A$, then the set of three teams $\{A, B, C\}$ is called "cyclic." Find the maximum and minimum number of cyclic sets. ${ }^{[1]}$
(2007, Italian National Team Selection Exam)
The maximum value is $\frac{n(n+1)(2 n+1)}{6}$, and the minimum value is 0.

Since any two teams must play a match, the total number of matches is $\mathrm{C}_{2 n+1}^{2}$.

To obtain the minimum number of cycles, let the number of wins for each team be $0,1, \cdots, 2 n$. In this case, there are no cyclic sets, i.e., the minimum value is 0.

In a non-cyclic set of three teams, there must be one team that has defeated the other two. Therefore, the number of non-cyclic sets containing $T_{i}(i=1,2, \cdots, 2 n+1)$ is $\mathrm{C}_{\mathrm{w}_{i}}^{2}$ (when $w_{i}=0$ or 1, it is stipulated that $\mathrm{C}_{\mathrm{w}_{i}}^{2}=0$).
Thus, the total number of all non-cyclic sets is $\sum_{i=1}^{2 n+1} C_{w_{i}}^{2}$.
$$
\begin{array}{l}
\text { and } \sum_{i=1}^{2 n+1} C_{w_{i}}^{2} \\
\geqslant \frac{1}{2(2 n+1)}\left(\sum_{i=1}^{2 n+1} w_{i}\right)^{2}-\frac{1}{2} \sum_{i=1}^{2 n+1} w_{i} \\
=\frac{n(n-1)(2 n+1)}{2} .
\end{array}
$$

Therefore, the number of cyclic sets is at most
$$
\mathrm{C}_{2 n+1}^{3}-\frac{n(n-1)(2 n+1)}{2}=\frac{n(n+1)(2 n+1)}{6} \text {, }
$$

which is achieved when and only when $w_{i}(i=1,2, \cdots, 2 n+1)$ are equal, reaching the maximum value. In this case, $w_{i}=n$.