Solution:
We can rewrite the given equation into
$$(x - 2007)(y - 2007) = 2007^{2} = 3^{4} \cdot 223^{2}$$
Since $x < y$, we have $x - 2007 < y - 2007$. It follows that
$$-2007 < x - 2007 < 2007 \quad \text{or} \quad |x - 2007| < 2007$$
Thus, we have $|y - 2007| > 2007$.
$$\begin{array}{rl} x - 2007 & y - 2007 \\ \hline 1 & 3^{4} \cdot 223^{2} \\ 3 & 3^{3} \cdot 223^{2} \\ 3^{2} & 3^{2} \cdot 223^{2} \\ 3^{3} & 3 \cdot 223^{2} \\ 3^{4} & 223^{2} \\ 223 & 3^{4} \cdot 223 \\ 3 \cdot 223 & 3^{3} \cdot 223 \end{array}$$
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.