jmc

Simplifying $\frac{1}{a}+\frac{1}{b}=\frac{2}{17}$, we have: $$\begin{array}{rl}2ab-17a-17b& =0\\ \Rightarrow 4ab-34a-34b+289& =289\\ \Rightarrow (2a-17)(2b-17)& =289.\end{array}$$Since $289=17^2$, we have three possibilities:

$2a-17=289$, $2b-17=1$ $2a-17=1$, $2b-17=289$ $2a-17=17$, $2b-17=17$

The first possibility gives us $a=153$, $b=9$, the second gives us $a=9$, $b=153$, and the last gives $a=b=17$. So, there are $\boxed{3}$ pairs of integers that satisfy the problem.