24. Determine all positive integers $n$ such that the equation $x^{3}+y^{3}+z^{3}=n x^{2} y^{2} z^{2}$ has positive integer solutions $(x, y, z)$.
24. Solution: By the symmetry of the equation, we may assume $x \geqslant y \geqslant z>0$. Transform the equation to: $n=\frac{x}{y^{2} z^{2}}+\frac{y}{x^{2} z^{2}}+\frac{z}{y^{2} x^{2}}$.

On the other hand, from the original equation, we get: $y^{3}+z^{3}=x^{2}\left(n y^{2} z^{2}-x\right)$, and $n y^{2} z^{2}-x$ is a positive integer, so $n y^{2} z^{2}-x \geqslant 1$.
Thus, from (2) we get $y^{3}+z^{3} \geqslant x^{2}$, i.e., $x \leqslant \sqrt{y^{3}+z^{3}}, x \geqslant y \geqslant z>0$. Substituting into (1) yields
$$
n \leqslant \frac{\sqrt{y^{3}+z^{3}}}{y^{2} z^{2}}+\frac{y}{y^{2} z^{2}}+\frac{z}{y^{2} z^{2}}=\sqrt{\frac{1}{y^{1} z}+\frac{1}{y z^{4}}}+\frac{1}{y z^{2}}+\frac{1}{y^{2} z} .
$$
(1) When $z \geqslant 2$, since $y \geqslant z$, from (3) we get $n \leqslant \frac{1}{2}$, which contradicts $n$ being a positive integer!
(2) When $z=1$, we get $y \geqslant z=1$;
(i) If $y=1$, from $x^{2} \leqslant y^{3}+z^{3}=2$, we get $x=1$, in which case $n=3$, and the solution to the equation is $(x, y, z)=(1,1,1)$.
(ii) If $y \geqslant 2$, and $z=1$, from (3) we get $n \leqslant \frac{3}{2}$, so $n$ can only be 1. The original equation transforms to: $x^{3}+y^{3}+1=x^{2} y^{2}$. In this case, the positive integer solution is $(x, y, z)=(3,2,1)$.
In summary, the value of $n$ is 1 or 3.