The only functions are the identity, zero, and the two periodic functions $$ f(x) = \begin{cases} 0 & x\equiv0\pmod 2 \\ 2n & x\equiv 1\pmod 2 \end{cases} $$for arbitrary integer $n$ and $$ f(x) = \begin{cases} 0 & x\equiv0\pmod 4 \\ 4n_1 + 2 & x\equiv 1\pmod 4 \\ 0 & x\equiv 2\pmod 4 \\ 4n_3 + 2 & x\equiv 3\pmod 4 \end{cases} $$for arbitrary integers $n_1$, $n_2$, and $n_3$. To check that the latter two work, note in each case that $f\circ f \equiv 0$ and $f(y)-2y$ is divisible by the period for all integers $y$. We now prove these are the only ones. Firstly, if $f$ is injective, then setting $x=2f(y)$ gives $f(2f(y)-2y)=0$, so $2(f(y)-y)$ is constant. This implies that $f$ is linear, and it's straightforward to check that only the identity works. Otherwise, suppose $f(a)=f(b)$ for distinct integers $a>b$. Then by substituting $y=a$ and $y=b$ and comparing, we find $f$ is periodic with period $2a-2b$. This implies that the image of $f$ is finite, and thus has a maximal element, say $M$. Suppose $f(k)=M$. If $f(f(y_0)) > 0$ for some $y_0$, then $f(k+2y_0 - f(y_0) ) > M$, a contradiction. So $f(f(y)) \leq 0$ for all $y$. By considering instead the minimal element in the image, we similarly find that $f(f(y)) \geq 0$ for all $y$. So $f\circ f \equiv 0$, which implies $f(0)=0$. The functional equation then reduces to $$ f(x-f(y)) = f(x-2y).$$Let $ P = \{ c \in \mathbb Z : f(x)=f(x+c) \text{ for all } x \in \mathbb Z \} $ be the set of periods of $f$, and let $k$ be the minimal positive element of $P$. Observe that: from the above equation, $f(y)-2y \in P$ for each $y$. In particular, $2f(y) \in P$ for all $y$. $P$ is closed under integer linear combinations of its elements, and by Bézout, every element of $P$ is divisible by $k$. By the first observation $f(1) - 2 \in P$ and $2f(1) \in P$. Then by the second, $4 = 2f(1) - 2(f(1)-2) \in P$, so $k\mid 4$. By individually considering the $k=1,2,4$ cases and adhering to the requirement that $k\mid f(y) - 2y$, we obtain the functions described earlier. @below Fixed; thanks.

@above, if you take n to be 1 mod 4, then f(f(n)) isn't 0 in your mod 4 solution, unless n_2=0 The thing is, f(f(y))=0 and f(y)-2y being a period implies f(2y)=0, so f has to be 0 for 2 mod 4 too. Other than that, nice solution! :>

We uploaded our solution https://calimath.org/pdf/MalaysiaAPMOSelection2024-3.pdf on youtube https://youtu.be/FEmh7nBaXLs.