$11 \cdot 20$ On a plane, there are $h+s$ lines, where $h$ lines are horizontal, and $s$ lines satisfy:
(1) None of them are horizontal.
(2) No two of them are parallel.
(3) Any three lines among 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 $(h, s)$.
[Solution] Since a straight line divides the plane into two regions.
If $n$ lines divide the plane into $a_{n}$ regions, then $n+1$ lines divide the plane into $a_{n+1}$ regions, which satisfies
$$
\begin{aligned}
a_{n+1} & =a_{n}+(n+1), \\
a_{1} & =2 .
\end{aligned}
$$

From this, we derive
$$
a_{n}=a_{1}+(2+3+\cdots+n)=1+\frac{n(n+1)}{2} .
$$

Thus, $s$ lines divide the plane into $1+\frac{s(s+1)}{2}$ regions.
Additionally, $h$ parallel lines intersecting these $s$ lines divide the plane into $h(s+1)$ regions (i.e., each additional horizontal line adds $s+1$ regions). Therefore, we have
$$
\begin{array}{c}
h(s+1)+1+\frac{s(s+1)}{2}=1992 . \\
(s+1)(2 h+s)=2 \times 1991=2 \times 11 \times 181 .
\end{array}
$$

For the above indeterminate equation, we can list 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 181 & 180 & 22 & $<0$ \\
\hline
\end{tabular}

Therefore, the pairs of positive integers $(h, s)$ that satisfy the requirement are
$$
(995,1),(176,10),(80,21) \text {. }
$$