17. How many ordered pairs $(x, y)$ of positive integers, where $x<y$, satisfy the equation
$$
\frac{1}{x}+\frac{1}{y}=\frac{1}{2007} .
$$
17. seven
We can rewrite the given equation into
$$
(x-2007)(y-2007)=2007^{2}=3^{4} \cdot 223^{2} .
$$

Since $x2007$.
\begin{tabular}{rl}
$x-2007$ & $y-2007$ \\
\hline 1 & $3^{4} \cdot 223^{2}$ \\
3 & $3^{3} \cdot 223^{2}$ \\
$3^{2}$ & $3^{2} \cdot 223^{2}$ \\
$3^{3}$ & $3 \cdot 223^{2}$ \\
$3^{4}$ & $223^{2}$ \\
223 & $3^{4} \cdot 223$ \\
$3 \cdot 223$ & $3^{3} \cdot 223$
\end{tabular}

For every pair of values of $x-2007$ and $y-2007$ in the above table, there is a corresponding pair of $x$ and $y$. Thus, there are seven such ordered pairs.