Example 11 (2002 Western Mathematical Olympiad) Find all positive integers $n$, such that
$$
n^{4}-4 n^{3}+22 n^{2}-36 n+18
$$
We denote
$$
A=n^{4}-4 n^{3}+22 n^{2}-36 n+18=\left(n^{2}-2 n\right)^{2}+18\left(n^{2}-2 n\right)+18 \text {. }
$$

Let $n^{2}-2 n=x, A=y^{2}, y$ be a non-negative integer, then
$$
(x+9)^{2}-63=y^{2} \text {, i.e., }(x+9-y)(x+9+y)=63 \text {. }
$$

It can only be $(x+9-y, x+9+y)=(1,63),(3,21)$ or $(7,9)$. We get $(x, y)=(23,31)$, $(3,9)$ or $(-1,1)$. Among these, only when $x=3$ and -1, $n^{2}-2 n=x$ has positive integer solutions, yielding $n=1$ or 3.
Therefore, the $n$ that satisfies the condition is 1 or 3.
Note: This problem can also be handled using inequalities.