4・155 To find the minimum value of $n$ for which the following system of equations
$$\left\{\begin{array}{l}
\sin x_{1}+\sin x_{2}+\cdots+\sin x_{n}=0 \\
\sin x_{1}+2 \sin x_{2}+\cdots+n \sin x_{n}=100
\end{array}\right.$$

has a solution.
[Solution] If $n<20$, then we subtract ten times equation (1) from equation (2), obtaining $-9 \sin x_{1}-8 \sin x_{2}-\cdots-\sin x_{9}+\sin x_{11}+2 \sin x_{12}+\cdots+(n-$ 10) $\sin x_{n}$
$$=100 \text {. }$$
The absolute value of the left side of (3) does not exceed $(9+8+\cdots+1) \times 2=90$, so (3) cannot hold. Therefore, the original system of equations has no solution when $n<20$.

When $n=20$, we can choose $x_{1}, x_{2}, \cdots, x_{20}$ such that
$$\begin{array}{l}
\sin x_{i}=-1,(i=1,2, \cdots, 10) \\
\sin x_{j}=1,(j=11,12, \cdots, 20)
\end{array}$$

Such a choice of $x_{1}, x_{2}, \cdots, x_{20}$ is clearly a solution to the original system of equations when $n=20$. Therefore, the smallest value of $n$ for which the original system of equations has a solution is 20.