II. (50 points) Find all positive integers $n$ that can be expressed as $n=$ $\frac{(a+b+c)^{2}}{a b c}\left(a, b, c \in \mathbf{Z}_{+}\right)$.
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 the other root of equation (1) be $c^{\prime}$, then
$$
\begin{array}{l}
c+c^{\prime}=n a b-2(a+b) \in \mathbf{Z}, \\
x^{\prime}=(a+b)^{2}>0, \text { so } c^{\prime} \in \mathbf{Z}_{+} .
\end{array}
$$

Thus, $\left(a, b, c^{\prime}\right)$ is also a solution that satisfies the problem's conditions, and $c^{\prime} \geqslant c$. Therefore, $(a+b)^{2}=c^{\prime} \geqslant c^{2}$, hence $c \leqslant a+b$.
$$
\begin{array}{l} 
\text { Also, } \frac{1}{b} \leqslant \frac{1}{a}, \frac{a}{b} \leqslant 1, \frac{b}{c} \leqslant 1, \text { so } \\
n=\frac{2}{a}+\frac{2}{b}+\frac{2}{c}+\frac{a}{c b}+\frac{b}{a c}+\frac{c}{a b} \\
\leqslant \frac{2}{a}+\frac{2}{b}+\frac{2}{c}+\frac{1}{c}+\frac{1}{a}+\left(\frac{1}{a}+\frac{1}{b}\right) \\
\leqslant \frac{3}{c}+\frac{7}{a} .
\end{array}
$$
(1) When $c=1$, $a=b=1$, then $n=9$.
(2) When $c \geqslant 2$, $n \leqslant \frac{3}{2}+7<9$, then $n \leqslant 8$.
Next, we prove that $n \neq 7$. Assume $n=7$.
If $a \geqslant 2$, then $c \geqslant 2,7=n \leqslant \frac{3}{2}+\frac{7}{2}=5$, which is a contradiction.
If $a=1$, then $1 \leqslant b \leqslant c \leqslant 1+b$.
So, $c=b$ or $1+b$.
When $c=b$, $(2 b+1)^{2}=7 b^{2}$, $b$ has no integer solution;
When $c=1+b$, $(2 b+2)^{2}=7 b(1+b)$,
which simplifies to $4(b+1)=7 b$, $b$ has no integer solution.
Therefore, $n \neq 7$.
(3) Let $F(a, b, c)=\frac{(a+b+c)^{2}}{a b c}$.
Since $F(k, k, k)=\frac{9}{k}$,
then when $k=1,3,9$, $n=9,3,1$.

Also, $F\left(k^{\prime}, k^{\prime}, 2 k^{\prime}\right)=\frac{8}{k^{\prime}}$,
then when $k^{\prime}=1,2,4$, $n=8,4,2$.
Clearly, $F(1,2,3)=6, F(1,4,5)=5$.
Thus, the positive integers $n$ that satisfy the problem's conditions are $1,2,3,4,5,6,8,9$.