First, let $a$, $b$ and $c$ be even: $a = 2a_1$, $b = 2b_1$, $c = 2c_1$. Then the number
$$a^2 + b^2 + c^2 = 4(a_1^2 + b_1^2 + c_1^2)$$
is divisible by $4$.

Now, let us prove the converse. Assume that $a^2 + b^2 + c^2$ is divisible by $4$. If exactly one of the numbers $a$, $b$ and $c$ were odd or if all three were odd, then the sum $a^2 + b^2 + c^2$ would be odd. This is not the case.

Finally, assume that exactly two of the numbers $a$, $b$ and $c$ are odd. We may assume that $a$ and $b$ are odd and $c$ is even. Let $a = 2a_1 - 1$, $b = 2b_1 - 1$ and $c = 2c_1$ for some positive integers $a_1$, $b_1$ and $c_1$. In this case the number
$$a^2 + b^2 + c^2 = (2a_1 - 1)^2 + (2b_1 - 1)^2 + (2c_1)^2 = 4(a_1^2 - a_1 + b_1^2 - b_1 + c_1^2) - 2$$
is not divisible by $4$. This contradicts the assumption. The only remaining possibility is that all three numbers are even.