Example 10 Let $p, q$ be given positive integers, $n=2^{p} \cdot 3^{q}$, find the number of positive integers less than $n$ and not divisors of $n$ among the positive divisors of $n^{2}$ (when $p=31, q=19$, this problem is from the 13th American Invitational Mathematics Examination)
Solve: Since the positive divisors of $n^{2}$ are of the form: $d=2^{\alpha} \cdot 3^{\beta}(1 \leqslant \alpha \leqslant 2 p, 0 \leqslant \beta \leqslant 2 q)$, we need to find all $d$ such that $0 \leqslant \alpha \leqslant p, 0 \leqslant \beta \leqslant q$ do not hold simultaneously, and $d$ is less than $n$.
$$
\begin{array}{c}
\text { Let } X=\left\{2^{\alpha} \cdot 3^{\beta} \mid p<\alpha \leqslant 2 p, 0 \leqslant \beta<q\right\}, \\
Y=\left\{2^{\alpha} \cdot 3^{\beta} \mid 0 \leqslant \alpha<p, q<\beta \leqslant 2 q\right\} .
\end{array}
$$

Thus, the required $d \in X \cup Y$. Since $X \cap Y=\varnothing$, we establish the following mapping $f: X \rightarrow Y$, such that for any $x=2^{\alpha} \cdot 3^{\beta} \in X$, we have $y=f(x)=2^{2 p-\alpha} \cdot 3^{2 q-\beta} \in Y$. It is easy to prove that this mapping is bijective, and for this $f$, when $x \in X$, $x y=n f(x)=2^{2 p} \cdot 3^{2 q}$ and $x, y$ are not equal to $n$, thus $x, y$ each belong to $n$ exactly once. From the above discussion, the number of the required positive integers $d$ is
$$
\frac{1}{2}|X \cup Y|=\frac{1}{2}(|X|+|Y|)=\frac{1}{2}(p \times q+q \times p)=p q .
$$