4. Express 2008 as the sum of $k\left(k \in \mathbf{N}_{+}\right)$ distinct square numbers. Then the minimum value of $k$ is ( ).
(A) 2
(B) 3
(C) 4
(D) 5
4.B.

Since the square of an odd number is congruent to 1 modulo 8, and the square of an even number is congruent to 0 modulo 4, if 2008 is the sum of two squares, i.e., $a^{2}+b^{2}=2008$, then $a$ and $b$ must both be even. Let $a=2a_{1}, b=2b_{1}$, transforming it into $a_{1}^{2}+b_{1}^{2}=502$, then $a_{1}$ and $b_{1}$ must both be odd. However, 502 is congruent to 6 modulo 8, which is a contradiction.
Therefore, $k \geqslant 3$.
When $k=3$, suppose $a^{2}+b^{2}+c^{2}=2008$, then $a$, $b$, and $c$ must all be even. Let $a=2a_{1}, b=2b_{1}, c=2c_{1}$, transforming it into $a_{1}^{2}+b_{1}^{2}+c_{1}^{2}=502$. The solution is
$\left(a_{1}, b_{1}, c_{1}\right)=(5,6,21)$.
Thus, $10^{2}+12^{2}+42^{2}=2008$.
Therefore, the minimum value of $k$ is 3.
Note: Since the squares are required to be distinct, option (D) does not contradict Lagrange's four-square theorem.