Answer: 71

Solution:

Using $k$ stamps of the denominations $n$ and $n + 1$ produces exactly the amounts $kn + c$ for $0 \le c \le k,$ and adding $5$-cent stamps then covers everything above in the same residue class mod $5.$ So in each class $r$ every amount at least $m(r)$ is formable and nothing smaller is, where $m(r)$ is the least value of $kn + c$ $(0 \le c \le k)$ congruent to $r$ mod $5.$ The greatest non-formable amount is $\max_r m(r) - 5,$ so we need $\max_r m(r) = 96:$ the class of $96$ (which is $1$ mod $5$) must be covered first exactly at $96,$ and every other class no later.

Case on $n \bmod 5,$ noting $kn + c \equiv kn + c.$ If $n \equiv 4:$ class $1$ needs $4k + c \equiv 1$ with $c \le k,$ first possible at $k = 4,$ $c = 0,$ so $4n = 96$ and $n = 24;$ the other classes are covered at $24,$ $48,$ $72,$ all less than $96,$ so $n = 24$ works. If $n \equiv 2:$ class $1$ is first covered at $k = 2,$ $c = 2,$ so $2n + 2 = 96$ and $n = 47;$ the other classes are covered at $47,$ $48,$ $94 \le 96,$ so $n = 47$ works.

If $n \equiv 3:$ class $1$ first at $2n = 96,$ so $n = 48,$ but then class $2$ is first covered at $2n + 1 = 97 \gt 96$ — fails. If $n \equiv 1:$ class $1$ first at $n = 96,$ but then class $3$ is first covered at $2n + 1 = 193$ — fails. If $n \equiv 0:$ class $1$ first at $n + 1 = 96,$ so $n = 95,$ but class $4$ needs $c = 4,$ $k \ge 4,$ giving $4n + 4 \gt 96$ — fails. The answer is $24 + 47 = 71.$