, 2018

In given condition, substitute $m=n$, we have
$$\max \{2 f(n), 2 n\} \mid \min \{4 n, f(2 n)+1\},$$
thus $4 n \geq 2 f(n)$ or $f(n) \leq 2 n$.
Continue substitute $m=1$, we have
$$\max \{2+f(n), 1+n\} \mid \min \{2+2 n, f(1+n)+1\} \quad (*)$$
Thus $f(n+1)+1 \geq f(n)+2$ or $f(n+1) \geq f(n)+1$. Since $f(1)=2$, we have $f(n) \geq n+1, \forall n \in \mathbb{Z}^{+}$.
From these, we can conclude that $f(2) \in \{3 ; 4\}$. But we have $f(2) \neq 4$ then $f(2)=3$. We shall prove by induction that $f(n)=n+1, \forall n \geq 2$.
The conclusion is true for $n=1,2$. Suppose that $f(n)=n+1$ for some $n \geq 2$.
Back to (*), note that $\max \{2+f(n), 1+n\}=f(n)+2=n+3$ and $\frac{2 n+2}{n+3} \in (1; 2)$ which implies that $n+3 \nmid 2 n+2$. Hence
$$\min \{2+2 n, f(1+n)+1\}=f(n+1)+1$$
Since $n+2 \leq f(n+1) \leq 2 n+2$, we must have $\frac{f(n+1)+1}{n+3}=1 \Leftrightarrow f(n+1)=n+2$. By induction hypothesis, we get $f(n)=n+1$.
Check condition, for all $m, n \in \mathbb{Z}^{+}$ then
$$\begin{gathered} \min \{2 m+2 n, f(m+n)+1\}=\min \{2 m+2 n, m+n+2\}=m+n+2 \text{ and } \\ \max \{f(m)+f(n), m+n\}=\max \{m+1+n+1, m+n\}=m+n+2 \end{gathered}$$
Hence, the function $f(n)=n+1$ satisfies the given condition.