1. $a, b$ are positive integers, satisfying: $\frac{1}{a}-\frac{1}{b}=\frac{1}{2018}$, then the number of all positive integer pairs $(a, b)$ is $\qquad$
Answer 4.
Analysis From $\frac{1}{a}-\frac{1}{b}=\frac{1}{2018}$, we know $1 \leq a2018$, since 1009 is a prime number, the possible factorizations of the number $2^{2} \cdot 1009^{2}$ are:
$$
1 \times 2018^{2}, 2 \times(2 \cdot 1009^{2}), 4 \times 1009^{2}, 1009 \times(4 \cdot 1009)
$$

Each of these factorizations corresponds to a solution for $(a, b)$, hence the number of solutions is 4.