［Solution］The given equation can be transformed into
$$
\begin{array}{c}
1991 y + 1991(x+1) + 1991 = (x+1) y, \\
(x+1) y - 1991(x+1) - 1991 y = 1991, \\
(x+1) y - 1991(x+1) - 1991 y + 1991^{2} = 1991 + 1991^{2}. \\
{[(x+1)-1991](y-1991) = 1991 \cdot 1992.}
\end{array}
$$

Obviously, the number of positive integer solutions to the given equation is equivalent to the number of positive divisors of $1991 \cdot 1992$, since
$$
1991 \cdot 1992 = 2^{3} \cdot 3 \cdot 11 \cdot 83 \cdot 181.
$$

Therefore, $1991 \cdot 1992$ has
$$
(3+1)(1+1)(1+1)(1+1)(1+1) = 64
$$

positive divisors, meaning the given equation has 64 sets of positive integer solutions.
4.5 Try to find the number of positive integer solutions to the indeterminate equation $\frac{1}{x+1}+\frac{1}{y}+\frac{1}{(x+1) y}=\frac{1}{1991}$.
(Japan Mathematical Olympiad, 1991)