4. The number of positive integer solutions to the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{1997}$ is
(A) 1
(B) 2
(C) 3
(D) greater than or equal to 4
4. (C).

It is easy to know that $x>1997, y>1997$. Let $x=1997+t_{1}$, $y=1997+t_{2}\left(t_{1}, t_{2}\right.$ be natural numbers $)$, then we have
$$
\frac{1}{1997+t_{1}}+\frac{1}{1997+t_{2}}=\frac{1}{1997} .
$$

Simplifying, we get $t_{1} t_{2}=1997^{2}$.
$\because t_{1}, t_{2}$ are natural numbers,
$$
\therefore t_{1} t_{2}=1 \times 1997^{2}=1997 \times 1997 \text {. }
$$

Thus, there are 2 corresponding sets of $x, y$ values. Since $x, y$ can be interchanged, there are a total of 3 sets of positive integers.