Bulgaria competition problems, 2022

$f(1) \in \{-80, -990\}$.
Since $f \in \mathbb{Z}[x]$, we have that $y - x \mid f(y) - f(x)$, $\forall x, y \in \mathbb{Z}$. Thus, $3 \mid |f(3) - f(0)|$, and $4 \mid |f(4) - f(0)|$. In the set $\{2, 20, 202, 2022\}$, only $2$ and $20$ are congruent modulo $3$, so $\{f(0), f(3)\} = \{2, 20\}$. Modulo $4$, $20$ has a different residue than the other three, so $f(0) \ne 20$. We conclude that $f(0) = 2$, $f(3) = 20$, while $f(4) = 202$ or $f(4) = 2022$.
Let $f(x) = ax^2 + bx + c$. From $c = f(0) = 2$ and $9a + 3b + 2 = f(3) = 20$, we deduce that $f(x) = ax^2 + (6 - 3a)x + 2$. Moreover $f(1) = 8 - 2a$, so $f \in \mathbb{Z}[x]$ whenever $a \in \mathbb{Z}$, due to
$$a = \frac{f(4) - f(0)}{4} - \frac{f(3) - f(0)}{3}.$$
Case 1: $f(4) = 202$. Then $a = \frac{202 - 2}{4} - \frac{20 - 2}{3} = \frac{200}{4} - \frac{18}{3} = 50 - 6 = 44$ and $f(1) = 8 - 2a = 8 - 88 = -80$.

Case 2: $f(4) = 2022$. Then $a = \frac{2022 - 2}{4} - \frac{20 - 2}{3} = \frac{2020}{4} - \frac{18}{3} = 505 - 6 = 499$ and $f(1) = 8 - 2a = 8 - 998 = -990$.