9. (22nd All-Soviet Union Mathematical Olympiad, 1988) Determine 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.
9. If $n<20$, then we subtract ten times equation (1) from equation (2), getting
$$
9 \sin x_{1}-8 \sin x_{2}-\cdots-\sin x_{9}+\sin x_{11}+2 \sin x_{12}+\cdots+(n-10) \sin x_{n}=100 \text {. }
$$
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 take $x_{1}, x_{2}, \cdots, x_{20}$ 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 $x_{1}, x_{2}, \cdots, x_{20}$ are clearly solutions 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.