Example 3.2.3 $2 n+1$ players $A_{1}, A_{2}, \cdots, A_{2 n+1}$ participate in a round-robin chess tournament, where each player plays one game against each of the other $2 n$ players, with no draws. Let $w_{i}$ be the number of games won by $A_{i}$. Find the minimum and maximum values of $S=\sum_{i=1}^{2 n+1} w_{i}^{2}$.
Solution: Since there are no ties, each match has one and only one winner, so $\sum_{i=1}^{2 n+1} w_{i}$ is equal to the number of matches,
$$
\sum_{i=1}^{2 n+1} w_{i}=\mathrm{C}_{2 n+1}^{2}=n(2 n+1) .
$$

By the Cauchy-Schwarz inequality $\left(\sum_{i=1}^{2 n+1} w_{i}^{2}\right)^{2 n+1} \sum_{i=1}^{2 n+1} 1^{2} \geqslant\left(\sum_{i=1}^{2 n+1} w_{i}\right)^{2}=n^{2}(2 n+1)^{2}$,
$$
S=\sum_{i=1}^{2 n+1} w_{i}^{2} \geqslant n^{2}(2 n+1) .
$$

For each $i, w_{i}=n$ when the equality holds. This situation is possible.
Suppose $P_{i} P_{j}$ play a match, when $i+j=$ odd, let the smaller number win, when $i+j=$ even, let the larger number win, then it is easy to prove $w_{i}=n(i=1,2, \cdots, 2 n+1)$,
$$
S_{\text {min }}=n^{2}(2 n+1) .
$$

To find $S_{\max }$, assume without loss of generality that $w_{1} \geqslant w_{2} \geqslant \cdots \geqslant w_{2 n+1}$,
If $w_{1}0$. If $w_{1}^{\prime}<2 n$, make the same adjustment, each adjustment can only increase $S$, and after adjustment, it must make $w_{1}=2 n$.

After the adjustment, the rearranged $w_{1}, w_{2}, \cdots, w_{2 n+1}$ can still be assumed to satisfy $w_{1} \geqslant w_{2} \geqslant \cdots \geqslant w_{2 n+1}$, where $w_{1}=2 n$. If $w_{2}<2 n-1$, make the same adjustment as above until $w_{2}=2 n-1$, and during the adjustment, $S$ always increases. Then adjust $w_{3}$ until $w_{3}=2 n-2 ; \cdots \cdots$; eventually we get
$$
\begin{array}{l}
S=\sum_{i=1}^{2 n+1} w_{i}^{2} \leqslant(2 n)^{2}+(2 n-1)^{2}+\cdots+2^{2}+1^{2}=\frac{1}{3} n(2 n+1)(4 n+1) . \\
w_{i}=2 n+1-i, \text { i.e., }\{2 n, 2 n-1,2 n-2, \cdots 2,1,0\}, \text { then } \sum_{i=1}^{2 n+1} w_{i}^{2}=\frac{n}{3}(2 n \\
+1)(4 n+1) . \\
\quad P_{1}, P_{2}, \cdots, P_{k}, \cdots, P_{2 n+1}
\end{array}
$$

It is sufficient for each $P_{k}$ to defeat all those who come after him to achieve the above scenario, so $S_{\max }=$
$$
\frac{n}{3}(2 n+1)(4 n+1) \text {. }
$$