jmc

By the Chicken McNugget theorem, the least possible value of $n$ such that $91$ cents cannot be formed satisfies $5n-(5+n)=91⟹n=24$, so $n$ must be at least $24$. For a value of $n$ to work, we must not only be unable to form the value $91$, but we must also be able to form the values $92$ through $96$, as with these five values, we can form any value greater than $96$ by using additional $5$ cent stamps. Notice that we must form the value $96$ without forming the value $91$. If we use any $5$ cent stamps when forming $96$, we could simply remove one to get $91$. This means that we must obtain the value $96$ using only stamps of denominations $n$ and $n+1$. Recalling that $n\ge 24$, we can easily figure out the working $(n,n+1)$ pairs that can used to obtain $96$, as we can use at most $\frac{96}{24}=4$ stamps without going over. The potential sets are $(24,25),(31,32),(32,33),(47,48),(48,49),(95,96)$, and $(96,97)$. The last two obviously do not work, since they are too large to form the values $92$ through $94$, and by a little testing, only $(24,25)$ and $(47,48)$ can form the necessary values, so $n\in \{24,47\}$. $24+47=\boxed{71}$.