1. Given that $p$ and $q$ are both natural numbers, and satisfy $19 p + 37 q = 1997$. Then $p q =$ $\qquad$
$=、 1.100$ or 60 .
From $19 p+97 q=1997$, we get
$$
p=\frac{1997-97 q}{19}=100+\frac{97(1-q)}{19} \text {. }
$$
Since $97$ and $19$ are coprime,
Therefore, $1-q$ is a multiple of 19.
$$
\begin{array}{l}
\text { Also } \frac{1997-97 q}{19} \geqslant 1, \\
\therefore q \leqslant 20 .
\end{array}
$$

Therefore, $1-q=0$ or $1-q=-19$.
Thus, $q=1$ or $q=20$.
At this time, the corresponding values of $p$ are 100 or 3.
$$
\therefore p q=100 \text { or } 60 \text {. }
$$