Find all $f : \mathbb{N} \to \mathbb{N}$ such that $f(a) + f(b) \mid a+b, \ \forall \ a, b \in \mathbb{N}$

As mentioned in the comments we have that $f(p-1) = p-1$ for a prime $p$. Now let $n \in \mathbb{N}$. There exists prime $p$ s.t. $p-1 > n$. Then we have:

$$f(n) + p - 1 \mid n + p-1$$

We have that the greatest proper divisor of $n+p-1$ is bounded from above by $\frac{n+p-1}{2} < p-1$

But $f(n) + p - 1 > p -1$, so it must be $n+p-1$. Hence $f(n) = n$ and the proof.