For a composite number $n$, let $f(n)$ denote the sum of its smallest three positive divisors, and $g(n)$ denote the sum of its largest two positive divisors. Find all composite numbers $n$ such that $g(n)$ is a positive integer power of $f(n)$.
II. Notice that when $n$ is odd, all divisors of $n$ are also odd.

Given that $f(n)$ is odd and $g(n)$ is even, it follows that $g(n)$ cannot be a positive integer power of $f(n)$. Therefore, we only need to consider the case where $n$ is even. In this case, 1 and 2 are the smallest two positive divisors of $n$, and $n$ and $\frac{n}{2}$ are the largest two positive divisors of $n$.
Let $d$ be the smallest positive divisor of $n$ other than 1 and 2.
If there exists $k \in \mathbf{Z}_{+}$ such that $g(n)=f^{k}(n)$, then
$$
\frac{3 n}{2}=(1+2+d)^{k}=(3+d)^{k} \equiv d^{k}(\bmod 3) \text {. }
$$

Since $\frac{3 n}{2}$ is clearly a multiple of 3, it follows that $3\left|d^{k} \Rightarrow 3\right| d$.
By the minimality of $d$, we have $d=3$.
Therefore, $\frac{3}{2} n=6^{k} \Rightarrow n=4 \times 6^{k-1}$.
Since $3 \mid n$, it follows that $k \geqslant 2$.
In summary, all possible values of $n$ are $n=4 \times 6^{l}\left(l \in \mathbf{Z}_{+}\right)$.