Problem 701

Given $x \geq y \geq z$, then
$$\begin{array}{l} 3 x^{3}=x^{3}+y^{3}+z^{3}=n x^{2} y^{2} z^{2} . \\ \therefore \quad 3 x=n y^{2} z^{3} . \\ \because \quad y, \quad z \in \mathbb{N}, \\ \therefore \quad 1+y^{3} z^{3} \geq y^{3}+z^{3} \\ \quad=n x^{2} y^{2} z^{2}-x^{3} \\ \quad=x^{2}\left(n y^{2} z^{2}-x\right) \geqslant x^{3} . \end{array}$$

From (1) and (2), we get
$$9\left(1+y^{3} z^{3}\right) \geqslant 9 x^{2} \geqslant n^{2} y^{4} z^{4} \text {. }$$
If $y z>1$, then $y^{4} z^{4}>1+y^{3} z^{3}$.
From (3), we get $9>-n^{2}$. Therefore, $n=1$ or 2. That is, $n \leqslant 4$.

Thus, it has been proven that (**) has no positive integer solutions. Now, we will discuss the cases for $n=1, 2, 3, 4$ separately.
For $n=1$, the solution is
$x=3 ; y=2, z=1$.
For $n=2$, (**) becomes
$$x^{3}+y^{3}+z^{3}=2 x^{2} y^{2} z^{2}.$$
From (1) and (3), we get $9\left(1+y^{3} z^{3}\right) \geqslant 4 y^{4} z^{4}$.
$\therefore \quad y z \leqslant 2$.
i) If $y z=1$, then (4) becomes
$$x^{3}+2=2 x^{2}.$$

This equation has no integer solutions, as $2 x^{2}-x^{3}$ is a multiple of 4, which cannot equal 2, leading to a contradiction. Therefore, equation (4) has no solutions.
ii) If $y z=2$, then (4) becomes
$$x^{3}+9=8 x^{2}.$$

This simplifies to $(x+1)\left(x^{2}-9 x+9\right)=0$.
This equation has no positive integer solutions, so (4) has no solutions.
For $n=3$, (**) has the solution
$$x=y=z=1.$$
For $n=4$, (**) becomes
$$x^{3}+y^{3}+z^{3}=4 x^{2} y^{2} z^{2}.$$

From (1), we know $y z=1$, so (5) becomes
$$x^{3}+2=4 x^{2}.$$

If this equation has a solution, then $x$ must be even. However, $4 x^{2}-x^{3}$ is a multiple of 4, leading to a contradiction. Therefore, (**) has no solutions in this case.

In summary, when $n=1$ or 3, (**) has positive integer solutions.