Find the sum of all positive integers $n$ such that, given an unlimited supply of stamps of denominations $5,n,$ and $n+1$ cents, $91$ cents is the greatest postage that cannot be formed.
By the Chicken McNugget theorem, the least possible value of $n$ such that $91$ cents cannot be formed satisfies $5n - (5 + n) = 91 \implies n = 24$, 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 $n \geq 24$, 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 $95$. By a little testing, only $(24, 25)$ and $(47, 48)$ can form the necessary values, so $n \in \{24, 47\}$. $24 + 47 = \boxed{071}$.
~Revision by [emerald_block](https://artofproblemsolving.com/wiki/index.php/User:Emerald_block) ~Minor Revision by [Mathkiddie](https://artofproblemsolving.com/wiki/index.php/Mathkiddie)

Note on finding and testing potential pairs
In order to find potential $(n,n+1)$ pairs, we simply test all combinations of $n$ and $n+1$ that sum to less than $4n$ (so that $n\ge24$) to see if they produce an integer value of $n$ when their sum is set to $96$. Note that, since $96$ is divisible by $1$, $2$, $3$, and $4$, we must either use only $n$ or only $n+1$, as otherwise, the sum is guaranteed to not be divisible by one of the numbers $2$, $3$, and $4$.
$\begin{array}{c|c|c} \text{Combination} & \text{Sum} & n\text{-value} \\ \hline n,n,n,n & 4n & 24 \\ n+1,n+1,n+1 & 3n+3 & 31 \\ n,n,n & 3n & 32 \\ n+1,n+1 & 2n+2 & 47 \\ n,n & 2n & 48 \\ n+1 & n+1 & 95 \\ n & n & 96 \\ \end{array}$
To test whether a pair works, we simply check that, using the number $5$ and the two numbers in the pair, it is impossible to form a sum of $91$, and it is possible to form sums of $92$, $93$, and $94$. ($95$ can always be formed using only $5$s, and the pair is already able to form $96$ because that was how it was found.) We simply need to reach the residues $2$, $3$, and $4$$\pmod{5}$ using only $n$ and $n+1$ without going over the number we are trying to form, while being unable to do so with the residue $1$. As stated in the above solution, the last two pairs are clearly too large to work.
$\begin{array}{c|c|c|c|c} \text{Pair} & \text{Not }91 & 92 & 93 & 94 \\ \hline 24,25 & \checkmark & \checkmark & \checkmark & \checkmark \\ 31,32 & \times & \checkmark & \checkmark & \checkmark \\ 32,33 & \times & \checkmark & \checkmark & \checkmark \\ 47,48 & \checkmark & \checkmark & \checkmark & \checkmark \\ 48,49 & \checkmark & \times & \checkmark & \checkmark \\ \end{array}$
(Note that if a pair is unable to fulfill a single requirement, there is no need to check the rest.)
~[emerald_block](https://artofproblemsolving.com/wiki/index.php/User:Emerald_block)