XXXI Brazilian Math Olympiad

First suppose that $r(A,B,n)$ is increasing for $n>n_0$. For the sake of simplicity, let $\bar{A}=N\setminus A$ and $\bar{B}=N\setminus B$. Then $$\begin{gathered} n+1=r(N,N,n)=r(A,B,n)+r(\bar{A},B,n)+r(A,\bar{B},n)+r(\bar{A},\bar{B},n) \\ ⟺r(\bar{A},B,n)+r(A,\bar{B},n)+r(\bar{A},\bar{B},n)=n+1-r(A,B,n) \end{gathered}$$ Since $r(A,B,n)$ is increasing, there exists a constant $c$ such that $n+1-r(A,B,n)\le c$ for all $n$. In particular, for all $n$, $r(\bar{A},N,n)=r(\bar{A},B,n)+r(\bar{A},\bar{B},n)<c$, implying $|\bar{A}|<c$. In fact, if $\bar{A}=\{a_1,a_2,a_3,\dots ,a_l,\dots \}$, $a_1<a_2<a_3<\dots$, for $a_k<n<a_{k+1}$, we have $r(\bar{A},N,n)=k$. Therefore if $\bar{A}$ were infinite, $r(\bar{A},N,n)$ would be unbounded. Analogously $\bar{B}$ must be finite as well.

Conversely, suppose $\bar{A}$ and $\bar{B}$ are both finite. For all $n>\max (\bar{A})+\max (\bar{B})$, if $a\in \bar{A}$ then $n-a>\max \bar{B}$, so $n-a\notin \bar{B}$, hence $n-a\in B$. In a similar fashion, if $b\in \bar{B}$ then $n-b\notin \bar{A}$, hence $n-b\in A$. Thus $$r(A,B,n)=n+1-|\bar{A}|-|\bar{B}|<n+2-|\bar{A}|-|\bar{B}|=r(A,B,n+1)$$ So $r(A,B,n)$ is increasing for $n>\max (\bar{A})+\max (\bar{B})$.