Saudi Arabian Mathematical Competitions

Taking $n=1$, we obtain $f(f(m)+2)=f(f(m))+2$, for all $m \in \mathbb{N}$. Taking $m=1$, we obtain $f(2+f(n))=4+f(n)$, for all $n \in \mathbb{N}$. Therefore
$$f(f(n))=f(n)+2, \quad \text{for all } n \in \mathbb{N} .$$
Because $f(2)=f(f(1))=f(1)+2=4$, we prove by induction that
$$f(n)=n+2$$
for all positive even integer $n \geq 2$.
Because the set $\{f(1)\} \cup \{f(2), f(4), f(6), \ldots\}$ contains all even positive integers and because $f$ is an injective function, the image by $f$ of any odd number $n \geq 3$ is an odd number.
Let
$$k=\min \{f(n) \mid f(n) \text{ is an odd integer }\}$$
and let $f\left(n_{0}\right)=k$ for some odd positive number $n_{0} \geq 3$. We prove by induction that
$$f(n)=n+2, \quad \text{for any odd number } n \geq k .$$
Indeed, $f(k)=f\left(f\left(n_{0}\right)\right)=f\left(n_{0}\right)+2=k+2$. Assume that $f(n)=n+2$, for some odd number $n \geq k$. We deduce that $f(n+2)=f(f(n))=f(n)+2= (n+2)+2$.
Clearly, $n_{0}<k$, otherwise $k=f\left(n_{0}\right)=n_{0}+2 \geq k+2$, which is impossible. Therefore,
$$\left\{f\left(n_{0}\right), f(k), f(k+2), f(k+4), \ldots\right\}=\{k, k+2, k+4, k+6, \ldots\}$$
covers all possible odd values taken by $f$. Because, $f$ is injective, and $f(3), f(5) \geq k$ are odd integers, we deduce that $n_{0}=3, k=5$ and
$$f(n)=n+2$$
for all positive odd integer $n \geq 2$.