Find all positive integers $n$ that can be expressed as
$$
n=\frac{(a+b+c)^{2}}{a b c} \quad\left(a, b, c \in \mathbf{N}^{*}\right)
$$
Let's assume $a \leqslant b \leqslant c$, and $(a, b, c)$ is the solution that minimizes $c$. From $n=\frac{(a+b+c)^{2}}{a b c}$, we get
$$
c^{2}+2(a+b) c+(a+b)^{2}=n a b c,
$$

which simplifies to
$$
c^{2}-(n a b-2(a+b)) c+(a+b)^{2}=0 .
$$

Let $c^{\prime}$ be the other root of equation (1). Then $c+c^{\prime}=n a b-2(a+b) \in \mathbf{Z}$, and $c c^{\prime}=(a+b)^{2}>0$, so $c^{\prime} \in \mathbf{N}^{*}$. Therefore, $(a, b, c^{\prime})$ is also a solution that satisfies the problem's conditions, and $c^{\prime} \geqslant c$. Thus, $(a+b)^{2}=c c^{\prime} \geqslant c^{2}$, implying $c \leqslant a+b$. Also, $\frac{1}{b} \leqslant \frac{1}{a}$, $\frac{a}{b} \leqslant 1$, and $\frac{b}{c} \leqslant 1$, so
$$
\begin{aligned}
n & =\frac{2}{a}+\frac{2}{b}+\frac{2}{c}+\frac{a}{c b}+\frac{b}{a c}+\frac{c}{a b} \\
& \leqslant \frac{2}{a}+\frac{2}{b}+\frac{2}{c}+\frac{1}{c}+\frac{1}{a}+\left(\frac{1}{a}+\frac{1}{b}\right) \leqslant \frac{3}{c}+\frac{7}{a}
\end{aligned}
$$
(1) When $c=1$, $a=b=1$, then $n=9$.
(2) When $c \geqslant 2$, $n \leqslant \frac{3}{2}+7<9$, so $n \leqslant 8$.

Next, we prove that $n \neq 7$. Assume $n=7$.
If $a \geqslant 2$, then $c \geqslant 2$, and $7=n \leqslant \frac{3}{2}+\frac{7}{2}=5$, which is a contradiction.
If $a=1$, then $1 \leqslant b \leqslant c \leqslant 1+b$. Therefore, $c=b$ or $1+b$.
When $c=b$, $(2 b+1)^{2}=7 b^{2}$, and $b$ has no integer solutions; when $c=1+b$, $(2 b+2)^{2}=7 b(1+b)$, which simplifies to $4(b+1)=7 b$, and $b$ has no integer solutions.
Thus, $n \neq 7$.
(3) Let $F(a, b, c)=\frac{(a+b+c)^{2}}{a b c}$. Since $F(k, k, k)=\frac{9}{k}$, when $k=1,3,9$, $n=9,3,1$.
Also, $F\left(k^{\prime}, k^{\prime}, 2 k^{\prime}\right)=\frac{8}{k^{\prime}}$, so when $k^{\prime}=1,2,4$, $n=8,4,2$. Clearly, $F(1,2,3)=6$ and $F(1,4,5)=5$. Therefore, the positive integers $n$ that satisfy the problem's conditions are $1,2,3,4,5,6,8,9$.