Team selection tests for IMO 2018

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