, 2023

From the assumptions of the problem, we have the following facts:
1. $f$ is injective
2. $f(m+n) \le f(m) + f(n)$
3. $|f(m+1) - f(m)| \le f(1)$
4. $(f(m+1) - f(1)) \mid f(m)$
Since $f$ is injective, $f(m+1) > 3f(1)$ for sufficiently large $m$. Combine the above listed facts, we have
$$2f(1) < f(m+1) - f(1) \mid |f(m) - f(m+1) + f(1)| \le 2f(1)$$
Therefore, we must have $f(m) - f(m+1) + f(1) = 0$. That is, $f(m) = mf(1) + c$ for some constant $c$ for sufficiently large $m$. Hence, for every sufficiently large $m, n$, we have $f(m+n) - f(m) = nf(1)$ divides $f(n) = nf(1) + c$, $c$ is forced to be 0. Claim, that $f(n) = nf(1)$ for all $n \in \mathbb{N}$. Toward contradiction, assume $n$ to be the largest number such that $f(n) \ne nf(1)$. Fix a sufficiently large $m$, we have $nf(1) = f(m+n) - f(m) \mid f(n)$, thus $f(n) = knf(1) = f(kn)$ for some $k \ge 2$, contradict to the fact that $f$ is injective.