3. The number of positive integer pairs $(x, y)$ that satisfy the equation
$$
\begin{array}{l}
x \sqrt{y}+y \sqrt{x}-\sqrt{2006 x}-\sqrt{2006 y}+\sqrt{2006 x y} \\
\quad=2006
\end{array}
$$
3.8 .

Given the equation can be transformed into
$$
(\sqrt{x}+\sqrt{y}+\sqrt{2006})(\sqrt{x y}-\sqrt{2006})=0 \text {. }
$$

Since $\sqrt{x}+\sqrt{y}+\sqrt{2006}>0$, we have
$$
\sqrt{x y}=\sqrt{2006} \text {, }
$$

which means $\square$
$$
\begin{array}{l}
x y=2006=1 \times 2006=2 \times 1003 \\
=17 \times 118=34 \times 59 .
\end{array}
$$

Thus, $(x, y)=(1,2006),(2006,1),(2,1003)$, $(1003,2),(17,118),(118,17),(34,59),(59,34)$. Therefore, there are 8 pairs of positive integers $(x, y)$ that satisfy the given equation.