APMO

Let $a_{h,s}$ be the number of regions formed by $h$ horizontal lines and $s$ other lines as described in the problem. Let ${\mathcal{F}}_{h,s}$ be the union of the $h+s$ lines and pick any line $\ell$. If it intersects the other lines in $n$ (distinct!) points then $\ell$ is partitioned into $n-1$ line segments and $2$ rays, which delimit regions. Therefore if we remove $\ell$ the number of regions decreases by exactly $n-1+2=n+1$.

Then $a_{0,0}=1$ (no lines means there is only one region), and since every one of the $s$ lines intersects the other $s-1$ lines, $a_{0,s}=a_{0,s-1}+s$ for $s\ge 0$. Summing yields

$$a_{0,s}=s+(s-1)+\cdots +1+a_{0,0}=\frac{s^2+s+2}{2}.$$

Each horizontal line only intersects the $s$ non-horizontal lines, so $a_{h,s}=a_{h-1,s}+s+1$, which implies $$a_{h,s}=a_{0,s}+h(s+1)=\frac{s^2+s+2}{2}+h(s+1).$$

Our final task is solving

$$a_{h,s}=1992⟺\frac{s^2+s+2}{2}+h(s+1)=1992⟺(s+1)(s+2h)=2\cdot 1991=2\cdot 11\cdot 181.$$

The divisors of $2\cdot 1991$ are $1,2,11,22,181,362,1991,3982$. Since $s,h>0,2\le s+1<s+2h$, so the possibilities for $s+1$ can only be $2,11$ and $22$, yielding the following possibilities for $(h,s)$ : $$(995,1),(176,10),\text{ and }(80,21).$$