Find all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that $f(1)=2$ and for all $m, n \in \mathbb{Z}_{>0}$, $\min (2 m+2 n, f(m+n)+1)$ is divisible by $\max (f(m)+f(n), m+n)$.
Substituting $m=n$ gives that $\min (4 n, f(2 n)+1)$ is divisible by $\max (2 f(n), 2 n)$. A number at most equal to $4 n$ is thus divisible by a number at least equal to $2 f(n)$. It follows that $4 n \geq 2 f(n)$, so $f(n) \leq 2 n$ for all $n$.
Substituting $m=n=1$ gives that $\min (4, f(2)+1)$ is divisible by $\max (2 f(1), 2)=$ $\max (4,2)=4$. Therefore, $\min (4, f(2)+1)$ cannot be less than 4, so $f(2)+1 \geq 4$. But we have also seen that $f(2) \leq 2 \cdot 2=4$, so $f(2)=3$ or $f(2)=4$.
First, assume that $f(2)=3$. We prove by induction on $n$ that $f(n)=n+1$ for all $n$. Suppose this is true for $n=r-1$ for some $r \geq 3$ and substitute $m=1$ and $n=r-1$. Then we get that $\min (2 r, f(r)+1)$ is divisible by $\max (f(1)+f(r-1), r)=$ $\max (r+2, r)=r+2$. Since $\min (2 r, f(r)+1) \leq 2 r+2$, it follows that $f(r)+1=r+2$ or $f(r)=r+1$. This completes the induction. We thus obtain the candidate function $f(n)=n+1$. We immediately check whether this function satisfies the condition. For all $m, n \in \mathbb{N}$, we have $\min (2 m+2 n, f(m+n)+1)=\min (2 m+2 n, m+n+2)=m+n+2$ because $m, n \geq 1$, and $\max (f(m)+f(n), m+n)=\max (m+n+2, m+n)=m+n+2$. The first is divisible by the second (since they are equal), so this function satisfies the condition.
Now assume that $f(2)=4$. We prove by induction on $n$ that $f(n)=2 n$ for all $n$. Suppose this is true for $n=r-1$ for some $r \geq 3$. We prove that $f(r)=2 r$. First, substitute $m=1$ and $n=r-1$. Then we find that $\min (2 r, f(r)+1)$ is divisible by $\max (f(1)+f(r-1), 1+r-1)=\max (2 r, r)=2 r$. Therefore, $\min (2 r, f(r)+1)$ cannot be less than $2 r$, so $f(r) \geq 2 r-1$. But we also knew that $f(r) \leq 2 r$, so $f(r) \in\{2 r-1,2 r\}$. Suppose that $f(r)=2 r-1$. Then substitute $m=1$ and $n=r$. We find that $\min (2(r+1), f(r+1)+1)$ is divisible by $\max (f(1)+f(r), 1+r)=\max (2+2 r-1, r+1)=2 r+1$. Since $2 r+1 \nmid 2(r+1)$, $2(r+1)$ cannot be the minimum, so $f(r+1)+1<2 r+2$ and it must also be divisible by $2 r+1$, so $f(r+1)=2 r$. Also substitute $m=2$ and $n=r-1$. Then we find that $\min (2(r+1), f(r+1)+1)=\min (2 r+2,2 r+1)=2 r+1$ is divisible by $\max (f(2)+f(r-1), 1+r)=\max (4+2 r-2, r+1)=2 r+2$, a contradiction. We conclude that $f(r)=2 r$, completing the induction. We thus obtain the candidate function $f(n)=2 n$ for all $n$. We check whether this function satisfies the condition. For all $m, n$, we have $\min (2 m+2 n, f(m+n)+1)=\min (2 m+2 n, 2 m+2 n+1)=2 m+2 n$ and $\max (f(m)+f(n), m+n)=\max (2 m+2 n, m+n)=2 m+2 n$. The first is divisible by the second (since they are equal), so this function satisfies the condition.
We conclude that there are exactly two solutions: the function $f(n)=n+1$ for all $n$ and the function $f(n)=2 n$ for all $n$.