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
$$n+1 = r(N, N, n) = r(A, B, n) + r(\bar{A}, B, n) + r(A, \bar{B}, n) + r(\bar{A}, \bar{B}, n) \\ \iff r(\bar{A}, B, n) + r(A, \bar{B}, n) + r(\bar{A}, \bar{B}, n) = n+1 - r(A, B, n)$$
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 $n > \max(\bar{A}) + \max(\bar{B})$.