2. 2. $21 \star \star$ 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, what is the minimum value of $n$?
If $n<20$, then we subtract ten times equation (1) from equation (2), obtaining
$$
\begin{array}{l}
-9 \sin x_{1}-8 \sin x_{2}-\cdots-\sin x_{9}+\sin x_{11} \\
\quad+2 \sin x_{12}+\cdots+(n-10) \sin x_{n}=100 .
\end{array}
$$
The absolute value of the left side of equation (3) does not exceed $(9+8+\cdots+1) \times 2=90$, so equation (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
$$
\sin x_{i}=-1(i=1,2, \cdots, 10) ; \sin x_{j}=1(j=11,12, \cdots, 20) .
$$

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.