4. If $\sqrt{m^{2}+1203}$ is an integer, then the sum of all positive integers $m$ that satisfy the condition is ( ).
(A)401
(B) 800
(C) 601
(D) 1203
4.B.

Let $\sqrt{m^{2}+1203}=n\left(n \in \mathbf{N}_{+}\right)$. Then
$$
\begin{array}{l}
n^{2}-m^{2}=1203 \\
\Rightarrow(n+m)(n-m)=401 \times 3=1203 \times 1 .
\end{array}
$$

Therefore, $\left\{\begin{array}{l}n+m=401, \\ n-m=3\end{array}\right.$ or $\left\{\begin{array}{l}n+m=1203, \\ n-m=1 .\end{array}\right.$
Solving these, we get $m=199$ or $m=601$. Hence, the sum is 800.