4. The number of positive integer pairs $(x, y)$ that satisfy the equation $x \sqrt{y}+\sqrt{x} y-\sqrt{2003 x}-\sqrt{2003 y}+$ $\sqrt{2003 x y}=200 B$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
4.(B).

From the given equation, we have
$$
\begin{array}{l}
(\sqrt{x y}-\sqrt{2003})(\sqrt{x}+\sqrt{y}+\sqrt{2003})=0 \\
\text { and } \sqrt{x}+\sqrt{y}+\sqrt{2003}>0,
\end{array}
$$

Therefore, $\sqrt{x y}-\sqrt{2003}=0$. Hence $x y=2003$.
Since 2003 is a prime number, it must be the case that
$$
\left\{\begin{array} { l } 
{ x = 1 , } \\
{ y = 2 0 0 3 }
\end{array} \text { or } \left\{\begin{array}{l}
x=2003, \\
y=1 .
\end{array}\right.\right.
$$