10. If $\sin \frac{\pi}{9}+\sin \frac{2 \pi}{9}+\cdots+\sin \frac{n \pi}{9}=\frac{1}{2} \tan \frac{4 \pi}{9}$, then the smallest positive integer $n$ is $\qquad$
10.4.

Notice,
$$
\begin{array}{l}
2 \sin \frac{\pi}{18}\left(\sum_{k=1}^{n} \sin \frac{k \pi}{9}\right) \\
=\sum_{k=1}^{n}\left(\cos \frac{(2 k-1) \pi}{18}-\cos \frac{(2 k+1) \pi}{18}\right) \\
=\cos \frac{\pi}{18}-\cos \frac{(2 n+1) \pi}{18} \\
\Rightarrow \tan \frac{4 \pi}{9} \cdot \sin \frac{\pi}{18}=\cos \frac{\pi}{18}-\cos \frac{(2 n+1) \pi}{18} \\
\Rightarrow \cos \frac{\pi}{18}=\cos \frac{\pi}{18}-\cos \frac{(2 n+1) \pi}{18} \\
\Rightarrow \cos \frac{(2 n+1) \pi}{18}=0 \\
\Rightarrow n=9 k+4(k \in \mathbf{Z}) .
\end{array}
$$

Therefore, the smallest positive integer $n$ is 4.