Example 2 In the plane, there are $h+s$ lines, of which $h$ are horizontal lines, and the other $s$ lines satisfy:
(1) None of them are horizontal lines;
(2) No two of them are parallel;
(3) Any 3 of the $h+s$ lines do not intersect at the same point, and these $h+s$ lines exactly divide the plane into 1992 regions. Find all pairs of positive integers $(h, s)$.
(1992 Asia Pacific Mathematical Olympiad Problem)
Solving: Since each line divides the plane into two regions, let $n$ lines (where each pair intersects, but no three lines intersect at the same point) divide the plane into $a_{n}$ regions. Then, $n+1$ such lines divide the plane into $a_{n+1}$ regions, satisfying $a_{n+1} = a_{n} + (n + 1)$, with $a_{1} = 2$.
From this, we derive $a_{n} = a_{1} + \sum_{k=2}^{n} (a_{k} - a_{k-1}) = 2 + \sum_{k=2}^{n} k = 1 + \frac{n(n+1)}{2}$.
Thus, $s$ lines divide the plane into $1 + \frac{1}{2} s(s+1)$ regions. When $h$ parallel lines intersect these $s$ lines, they add $h(s+1)$ regions (i.e., each additional horizontal line adds $s+1$ regions), so we have:
\[
\begin{array}{l}
h(s+1) + 1 + \frac{1}{2} s(s+1) = 1992, \\
(s+1)(2 h + s) = 2 \times 1991 = 2 \times 11 \times 181.
\end{array}
\]

The possible positive integer solutions to the above indeterminate equation can be listed in the following table:
\begin{tabular}{|c|c|c|c|}
\hline $s+1$ & $s$ & $2 h + s$ & $h$ \\
\hline 2 & 1 & 1991 & 995 \\
\hline 11 & 10 & 362 & 176 \\
\hline 22 & 21 & 181 & 80 \\
\hline $\geqslant 181$ & $\geqslant 180$ & $\leqslant 22$ & $<0$ \\
\hline
\end{tabular}

Therefore, the required positive integer pairs $(h, s)$ are $(995, 1)$, $(176, 10)$, and $(80, 21)$.