Four. (20 points) Given that the three sides of $\triangle A B C$ are $a, b, c$, and they satisfy
$$
a b c=2(a-1)(b-1)(c-1) .
$$
(1) Does there exist a $\triangle A B C$ with all sides being integers? If so, find the side lengths; if not, explain the reason.
(2) If $a>1, b>1, c>1$, find the minimum perimeter of $\triangle A B C$.
(1) Suppose $a \geqslant b \geqslant c$. Clearly, $c \geqslant 2$.
If $c \geqslant 5$, then $\frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c} \leqslant \frac{1}{5}$.
From $a b c=2(a-1)(b-1)(c-1)$, we get
$$
\frac{1}{2}=\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right) \geqslant\left(\frac{4}{5}\right)^{3} \text {. }
$$

This is a contradiction.
Therefore, $c$ can only be 2, 3, or 4.
When $c=2$, $a b=(a-1)(b-1)$, which gives $a+b=1$.
Since $a \geqslant b \geqslant 2$, there is no solution.
When $c=3$, $3 a b=4(a-1)(b-1)$, which simplifies to $(a-4)(b-4)=12$.
Since $a \geqslant b \geqslant 3$, we have
$\left\{\begin{array}{l}a-4=12, \\ b-4=1\end{array}\right.$ or $\left\{\begin{array}{l}a-4=6, \\ b-4=2\end{array}\right.$ or $\left\{\begin{array}{l}a-4=4, \\ b-4=3 .\end{array}\right.$
Solving these, we get $\left\{\begin{array}{l}a=16, \\ b=5\end{array}\right.$ or $\left\{\begin{array}{l}a=10, \\ b=6\end{array}\right.$ or $\left\{\begin{array}{l}a=8, \\ b=7 .\end{array}\right.$
The only solution that forms a triangle is $a=8, b=7, c=3$.
When $c=4$, similarly, we get
$$
a=9, b=4 \text { or } a=6, b=5 \text {. }
$$

The only solution that forms a triangle is $a=6, b=5, c=4$.
Therefore, there exist triangles $\triangle A B C$ with integer side lengths, whose side lengths are $4,5,6$ or $3,7,8$.
(2) From $a b c=2(a-1)(b-1)(c-1)$, we get
$$
\begin{array}{l}
\frac{1}{2}=\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right) \\
\leqslant\left[\frac{\left(1-\frac{1}{a}\right)+\left(1-\frac{1}{b}\right)+\left(1-\frac{1}{c}\right)}{3}\right]^{3} . \\
\text { Hence } \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \leqslant 3-\frac{3}{\sqrt[3]{2}} . \\
\text { Also, }(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9, \text { so } \\
a+b+c \geqslant \frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \geqslant \frac{9}{3-\frac{3}{\sqrt[3]{2}}} \\
=\frac{3 \sqrt[3]{2}}{\sqrt[3]{2}-1} .
\end{array}
$$

Thus, the minimum perimeter of $\triangle A B C$ is $\frac{3 \sqrt[3]{2}}{\sqrt[3]{2}-1}$, which is achieved when $a=b=c=\frac{\sqrt[3]{2}}{\sqrt[3]{2}-1}$.