4. Among all lines passing through the point $(\sqrt{2001}, 0)$, the number of lines passing through two different rational points (points with both coordinates being rational numbers) is ( ).
(A)0 lines
(B)infinitely many lines
(C)at least 1 line
(D)exactly 1 line
4.(D).

Obviously, there are no rational points on the line $x=\sqrt{2001}$, so the slope of the line exists. Let the equation of the line be
$$
y=k(x-\sqrt{2001}) \text {. }
$$

If there exist two different rational points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, then $k=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ is a rational number. And $y_{1}=k\left(x_{1}-\sqrt{2001}\right)$, so $k \sqrt{2001}=k x_{1}-y_{1}$.
The right side of the equation is a rational number, thus, $k=0$.
That is, the line that satisfies the condition is the $x$-axis.