Find all $f : mathbb{N} to mathbb{N} $ such that $f(a) + f(b)$ divides $2(a + b – 1)$ for all $a, b in mathbb{N}$.

I think the answer is

$boxed{f(x)=2x-1, f(x)=1}.$

$$P(1,1) implies f(1)+f(1)| 2(1+1-1)=2implies f(1)=1$$

$$P(1,a)implies f(1)+f(a)|2(1+a-1)=2a implies f(a)+1|2a implies f(a) le 2a-1.$$

We have $$P(1,2)implies f(1)+f(2)|2(1+2-1)=4implies f(2)=3~~text{or}~~f(2)=1.$$

For $p$ an odd prime. We get,

$$P(1,p)implies f(p)+1|2p implies f(p)=1,p-1, 2p-1$$

Now if $f(p)=p-1 $ for some $p$ then $$P(p,p)implies f(p)+f(p)=2(p-1) | 2(2p-1)$$

Not possible.

This was my progress… Any hints or solutions? I would prefer if someone can send a sketch of hints( Rather than a full solution).

Thanks in advance.