Answer $B$
Since, $m$ and $n$ are odd numbers. So, they can be expressed as:
$m = 2x+1, x\in Z$, and $ n = 2y + 1, y\in Z$
Now,
$m^2 + n^2 = (2x+1)^2 + (2y+1)^2 = 4x^2+1+4x + 4y^2+1+4y = 4 \underbrace {(x^2+x+y^2+y)}_k + 2 = 4k+2$
So, $m^2+n^2$ can be expressed as $m^2 + n^2 = 4k+2$, where $k \in Z$
for $k =\pm0, \pm1, \pm2, \pm3\cdots$
So, values obtained are: $0, \pm6, \pm14, \pm18. \cdots$
Hence, It will never be divisible by $4$.
Therefore, option $B$ is the correct answer.