Solution:
First substitute $x = p$, to get $p \mid a p^{2} + b p + c$ so $p \mid c$. Therefore $c = p$.

Next substitute $x = 1$, to get $p \mid a + b + c$. Since $c = p$, this gives us $p \mid a + b$.

Finally substitute $x = p - 1$, to get $(p - 1)^{2} a - (p - 1) b + c \equiv 0$ (mod $p$). Thus $a - b + c \equiv 0$ (mod $p$), and since $c = p$, we get $p \mid a - b$.

Adding and subtracting $p \mid a + b$ and $p \mid a - b$ we get
$$p \mid (a + b) + (a - b) = 2a \qquad \text{and} \qquad p \mid (a + b) - (a - b) = 2b.$$
Now there are two cases: either $p = 2$ or $p$ is odd.

Case 1 $p = 2$

$2 \mid a - b$ means that either $a = b = 1$ or $a = b = 2$, i.e. the only two possibilities for $(a, b, c)$ are $(1, 1, 2)$ and $(2, 2, 2)$. So the only candidate polynomials are
$$x^{2} + x + 2 \qquad \mathrm{and} \qquad 2x^{2} + 2x + 2.$$
Now we check that these work. If $f(x) = x^{2} + x + 2 = 2 + x(x + 1)$ which is always even because both 2 and $x(x + 1)$ are even for all integers $x$. If $f(x) = 2x^{2} + 2x + 2 = 2(x^{2} + x + 1)$ which is always even because it has a factor of 2.

Case 2 $p$ is an odd prime:

Since $(2, p) = 1$, $p \mid a$ so $a = p$. Also note that $p \mid 2b$ by subtracting $a - b$ from $a + b$, so by a similar argument, $b = p$. Thus, $a + b + c = p + p + p = 3p$

Combining the two cases, $a + b + c = 3p$ for all $p$ and $a + b + c = 4$ when $p = 2$ are the only possible values of $a + b + c$.