jmc

Multiplying out all of the denominators, we get: $$\begin{array}{rl}104(n+k)& <195n<105(n+k)\\ 0& <91n-104k<n+k\end{array}$$ Since $91n-104k<n+k$, $k>\frac{6}{7}n$. Also, $0<91n-104k$, so $k<\frac{7n}{8}$. Thus, $48n<56k<49n$. $k$ is unique if it is within a maximum range of $112$, so $n=\boxed{112}$.