Here is the function, $$f:\mathbb{N}\rightarrow \mathbb{N}, f(f(m)+f(n))=m+n$$. This is a one to one function. But I cannot proceed further. I have tried to arrange the domains and their respective ranges, like,
$f(f(3)+f(1))=4 \land f(2f(2))=4$.
But cannot proceed further. Thanks in anticipation.
Taking $m=n$ in the assumption $f(f(m)+f(n))=m+n$, we get $$\tag{1} f(2f(n))=2n\mbox{ for all }n\in\mathbb{N}.$$ Now, if $f(m)=f(n)$, multiplying it by $2$ and then consider its value under $f$, we get $$f(2f(m))=f(2f(n)),$$ which implies that $2m=2n$ by $(1)$, which gives $m=n$. This proves that $f$ is injective.