Example 3.2.2 Find the minimum value of $n$ that satisfies $\sin x_{1}+\sin x_{2}+\cdots+\sin x_{n}=0$, $\sin x_{1}+2 \sin x_{2}+\cdots+n \sin x_{n}=100$.
Solution: First, discuss when $\sum_{i=1}^{n} \sin x_{i}=0$, under what conditions $S_{n}=\sum_{i=1}^{n} i \sin x_{i}$ is maximized.
(1) To maximize $S_{n}$, let $i_{1}=\min \left\{i_{k} ; \sin x_{i_{k}} \geqslant 0\right\}, j_{m}=\max \left\{j_{k} ; \sin x_{j_{k}}j_{m}$.
If $i_{1}0$, to maximize $S_{n}$, it is necessary that $j_{m}0$.
From (1) and (2) and $\sum_{k=1}^{n} \sin x_{k}=0$, we can derive:
When $n=2 k$,
$$
S_{n}\left(\sin x_{1}, \sin x_{2}, \cdots, \sin x_{2 k}\right) \leqslant S_{n}(\overbrace{-1,-1 \cdots-1}^{k}, \overbrace{1,1 \cdots 1}^{k}),
$$
When $n=2 k+1$,
$$
S_{n}\left(\sin x_{1}, \sin x_{2}, \cdots, \sin x_{2 k+1}\right) \leqslant S_{n}(\overbrace{-1,-1 \cdots-1}^{k}, 0, \overbrace{1,1 \cdots 1}^{k}) .
$$

Therefore, when $k<10$, $S_{2 k} \leqslant-(1+2+\cdots+k)+[(k+1)+(k+2)+\cdots+$
$$
\begin{array}{l}
2 k]=k^{2}<100, \\
\quad k \leqslant 10 \text { when, } S_{2 k-1} \leqslant-(1+2+\cdots+k-1)+[(k+1)+\cdots+(2 k-1)] \\
=k(k-1)<100,
\end{array}
$$
When $n<20$, $S_{n}<100$. When $n=20$, $S_{20}(\overbrace{-1,-1 \cdots-1}^{10}, \overbrace{1,1 \cdots 1}^{10})=-\sum_{i=1}^{10} i+$ $\sum_{i=11}^{20} i=100$, so the minimum value of $n$ is 20.