Problem 671

(1) Let's assume $m \geqslant n \geqslant p \geqslant q$.
Clearly, $q \geqslant 2$.
If $q \geqslant 3$, then
$$\begin{aligned} \frac{1}{m} & \leqslant \frac{1}{n} \leqslant \frac{1}{p} \leqslant \frac{1}{q} \leqslant \frac{1}{3} \\ \Rightarrow & \frac{1}{6}=\left(1-\frac{1}{m}\right)\left(1-\frac{1}{n}\right)\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right) \\ & \geqslant\left(\frac{2}{3}\right)^{4}>\frac{1}{6} \end{aligned}$$

Contradiction.
Thus, $q=2$.
Substituting $q=2$ into the given equation, we get
$$\begin{array}{l} m n p=3(m-1)(n-1)(p-1) . \\ \text { Clearly, } p \geqslant 2 . \\ \text { If } p \geqslant 4, \text { then } \\ \frac{1}{m} \leqslant \frac{1}{n} \leqslant \frac{1}{p} \leqslant \frac{1}{4} \\ \Rightarrow \frac{1}{3}=\left(1-\frac{1}{m}\right)\left(1-\frac{1}{n}\right)\left(1-\frac{1}{p}\right) \\ \quad \geqslant\left(\frac{3}{4}\right)^{3}>\frac{1}{3}, \end{array}$$

Clearly, $p \geqslant 2$.
If $p \geqslant 4$, then
$$\begin{aligned} \frac{1}{m} & \leqslant \frac{1}{n} \leqslant \frac{1}{p} \leqslant \frac{1}{4} \\ \Rightarrow & \frac{1}{3}=\left(1-\frac{1}{m}\right)\left(1-\frac{1}{n}\right)\left(1-\frac{1}{p}\right) \\ & \geqslant\left(\frac{3}{4}\right)^{3}>\frac{1}{3}, \end{aligned}$$

Contradiction.
Thus, $p=2$ or 3.
When $p=2$, we have
$$\begin{aligned} 2 m n & =3(m-1)(n-1) \\ \Rightarrow & (m-3)(n-3)=6 \times 1=3 \times 2 \\ & \Rightarrow(m, n)=(9,4),(6,5) . \end{aligned}$$

When $p=3$, we have
$$\begin{array}{l} m n=2(m-1)(n-1) \\ \Rightarrow(m-2)(n-2)=2 \times 1 \\ \Rightarrow(m, n)=(4,3) . \\ \text { Therefore, }(m, n, p, q) \\ =(9,4,2,2),(6,5,2,2),(4,3,3,2) . \end{array}$$
(2) Note that,
$$\begin{array}{l} \frac{1}{6}=\left(1-\frac{1}{m}\right)\left(1-\frac{1}{n}\right)\left(1-\frac{1}{p}\right)\left(1-\frac{1}{q}\right) \\ \leqslant\left[\frac{\left(1-\frac{1}{m}\right)+\left(1-\frac{1}{n}\right)+\left(1-\frac{1}{p}\right)+\left(1-\frac{1}{q}\right)}{4}\right] . \end{array}$$

That is,
$$\begin{array}{l} \frac{1}{m}+\frac{1}{n}+\frac{1}{p}+\frac{1}{q} \leqslant 4-\frac{4}{\sqrt[4]{6}} . \\ \text { Also, }(m+n+p+q)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}+\frac{1}{q}\right) \geqslant 16 \\ \Rightarrow m+n+p+q \\ \quad \geqslant \frac{16}{\frac{1}{m}+\frac{1}{n}+\frac{1}{p}+\frac{1}{q}} \geqslant \frac{16}{4-\frac{4}{4}}=\frac{4 \sqrt[4]{6}}{\sqrt[4]{6}-1} . \end{array}$$

Equality holds if and only if $m=n=p=q=\frac{\sqrt[4]{6}}{\sqrt[4]{6}-1}$.
Thus, $(m+n+p+q)_{\min }=\frac{4 \sqrt[4]{6}}{\sqrt[4]{6}-1}$.