Let $n$ be a positive integer and the coefficient of the $x^3$ term in the expansion of $(1+\frac{x}{n})^n$ be $\frac{1}{16}$. Find $n$.
The $x^3$ term should be $\binom{n}{3}\frac{1}{n^3}x^3$. Therefore, we have $$\binom{n}{3}\frac{1}{n^3}=\frac{1}{16}\implies n=4, \frac{5}{4}$$
Because $n$ is a positive integer, therefore the answer is $\boxed{4}$.