Four, (50 points) In a round-robin tournament with $2 n+1$ teams, each team plays exactly one match against every other team, and there are no ties. If three teams $A, B, C$ satisfy: $A$ beats $B, B$ beats $C, C$ beats $A$, then they form a “cyclic triplet”. Find:
(1) the minimum possible number of cyclic triplets;
(2) the maximum possible number of cyclic triplets.
(1) The minimum value is 0.
For two participating teams $T_{i}$ and $T_{j}$, if and only if $i>j$, $T_{i}$ defeats $T_{j}$, at which point the number of cyclic triples is minimized.
(2) Any three participating teams either form a cyclic triple or a "dominant" triple (i.e., one team defeats the other two). Let the former have $c$ groups and the latter have $d$ groups. Then
$$
c+d=\mathrm{C}_{2 n+1}^{3} \text {. }
$$

Assume a team $T_{i}$ defeats $x_{i}$ other teams. Then the winning group must be among $\mathrm{C}_{x}^{2}$ dominant triples.
Note that, the total number of matches is
$$
\sum_{i=1}^{2 n+1} x_{i}=\mathrm{C}_{2 n+1}^{2} \text {. }
$$

Therefore, $d=\sum_{i=1}^{2 n+1} \mathrm{C}_{x_{i}}^{2}=\frac{1}{2} \sum_{i=1}^{2 n+1} x_{i}^{2}-\frac{1}{2} \mathrm{C}_{2 n+1}^{2}$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(2 n+1) \sum_{i=1}^{2 n+1} x_{i}^{2} \geqslant\left(\sum_{i=1}^{2 n+1} x_{i}\right)^{2}=n^{2}(2 n+1)^{2} . \\
\text { Hence } c=\mathrm{C}_{2 n+1}^{3}-\sum_{i=1}^{2 n+1} \mathrm{C}_{x_{i}}^{2} \\
\leqslant \mathrm{C}_{2 n+1}^{3}-\frac{n^{2}(2 n+1)}{2}+\frac{1}{2} \mathrm{C}_{2 n+1}^{2} \\
=\frac{1}{6} n(n+1)(2 n+1) .
\end{array}
$$

Arrange all participating teams in a circle. For each team, the $n$ teams in the clockwise direction are defeated by it, and the $n$ teams in the counterclockwise direction defeat it, achieving the maximum value.
(Zhou Shundi, Hangzhou High School, Zhejiang Province, 310003)