Solution:

Suppose he asks $n$ questions as usual, and then asks "did you lie to any of the last $n$ questions?" If the reply is a truthful no, then the $n$ answers were correct. If the reply is a lying no, then the $n$ answers were still correct. On the other hand if the answer is yes, then the $n$ answers might have been correct and might not. However, a lie has certainly been told, so all future answers must be truthful and so he could ask the $n$ questions again.

$91 = 7 \times 13$, so the obvious candidates for $n$ are $7$ and $13$. If we take $n = 7$, then the worst case is $13$ check questions and $7$ repeat questions. That does not work because he needs $20$ extra questions and only has $14$. A little thought suggests reducing $n$ each time. So the first batch of questions is $13$, followed by a check question. If the check answer is yes, then he knows a lie has been told and asks the $13$ questions again. No further check questions are needed, and he has used exactly $14$ extra questions. If the check answer is no, then the lie may not have been told, so the next batch of questions is $12$, followed by a check question, and so on. That allows him to ask $13 + 12 + \ldots + 1 = 91$ questions. If he gets a yes to the check question after the batch of $i$, then he ignores the answers to that batch and asks them again, thus asking a total of $14$ extra questions, but thereafter asks no check questions.