30. Find all $n \in \mathbf{N}^{*}$, such that there exist $x, y, z \in \mathbf{N}^{*}$, satisfying:
$$n=\frac{(x+y+z)^{2}}{x y z}$$
30. Let the set of $n$ that meets the requirements be $S$. By taking $(x, y, z)=(9,9,9)$, $(4,4,8)$, $(3,3,3)$, $(2,2,4)$, $(1,4,5)$, $(1,2,3)$, $(1,1,2)$, $(1,1,1)$, we know $\{1,2,3,4,5,6,8,9\} \subseteq S$.

Now, let $n \in S$, and $x, y, z$ be the positive integers that satisfy $n=\frac{(x+y+z)^{2}}{x y z}$, such that $x+y+z$ is minimized.

We prove: $x \leqslant y+z$, $y \leqslant z+x$, $z \leqslant x+y$.
To prove (10) holds, by symmetry, we only need to prove: $x \leqslant y+z$.
If $x>y+z$, then by the condition, $x \mid (x+y+z)^{2}$, so $x \mid (y+z)^{2}$. Now let $x^{\prime}=\frac{(y+z)^{2}}{x}$, then $x^{\prime} \geq 1$. If $x^{\prime} \geq 1$, then $y z - (y+z) = (y-1)(z-1) - 1 \geq 0$, i.e., $y z \geq y + z \geq x$. At this point, combining $x \geq y \geq z$, we have
$$\begin{aligned}
n & =\frac{x^{2}+y^{2}+z^{2}+2 x y+2 y z+2 z x}{x y z}=2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{x}{y z}+\frac{y}{z x}+\frac{z}{x y} \\
& \leqslant 2\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\right)+1+1+1=6
\end{aligned}$$

Therefore, $n \in \{1,2,3,4,5,6,8,9\}$.
In summary, $S=\{1,2,3,4,5,6,8,9\}$, which is the set of all $n$ that meet the conditions.