How can we solve it?

The fundamental period of $f(x)=\frac{\sin(nx)}{\tan(\frac{x}{n})}$ is the least positive integer multiple of both the period of $\sin(nx)$ and the period of $\tan(\frac{x}{n})$ .

Note that the period of $\sin(nx)$ is $\frac{2\pi}{n}$ and the period of $\tan(\frac{x}{n})$ is $n\pi$ .

When $n=1$ , the period of $\sin(nx)$ is $2\pi$ and the period of $\tan(\frac{x}{n})$ is $\pi$ , so the period of $f(x)$ is $2\pi$ .

When $n=2$ , the period of $\sin(nx)$ is $\pi$ and the period of $\tan(\frac{x}{n})$ is $2\pi$ , so the period of $f(x)$ is $2\pi$ .

When $n=3$ , the period of $\sin(nx)$ is $\frac{2\pi}{3}$ and the period of $\tan(\frac{x}{n})$ is $3\pi$ , so the period of $f(x)$ is $6\pi$ .

When $n=4$ , the period of $\sin(nx)$ is $\frac{\pi}{2}$ and the period of $\tan(\frac{x}{n})$ is $4\pi$ , so the period of $f(x)$ is $4\pi$ .

Hence the answer is $\boxed{n=3}$ .