I found somewhere this 'knights and knaves' problem (I can't locate the source anymore):
An Island is inhabited by exactly $n$ people, with $n>2$, of whom some (the exact number is not known) are knights, who alwais tell the truth, and some are knaves (the exact number is not known, but it is known that there is at least 1 knave). Every knave always lies, except for one of them (exactly one), their boss, who always tells the truth. An explorer arrives to this Island, and he knows all the information reported in the text of this problem, the name of all the inhabitants of the island, and nothing else. He can ask any number of question to anyone, but every question must be of the kind "Is X a knight or a knave?". What is the least integer number $m$ so that the explorer has a strategy to be certain of guessing correctly who is the knaves' boss with no more than $m$ questions?
I showed that $m\le n$ if $n$ is even and $m\le n+1$ if $n$ is odd by the following simple argument: let $1,2,\dots n$ be the names of the islanders, one proceeds asking to 1 about 2 and viceversa; if both answer the same none of them is the knaves' boss, otherwise one of them is. So by doing this trick with couples $2k-1,2k$ for $k=1,2,\dots \lceil n/2\rceil-1$, we have surely found a group of 2 people at most of whom one is the knaves' boss. WLOG, let's assume he is one of 1 and 2. Now ask 1 about 3 and viceversa, and you will be able to guess with certainty who the boss is. This procedure leads to the $n$ / $n+1$ upper bound on $m$. But I can't find any way to try showing this is indeed the minimum (or find a better procedure). I'd be very glad if someone may help me on this.
You can ask $X$ a question "Is $X$ a knave?". Only the knave boss will answer "Yes" for this question, thus there is possibility to ask $n-1$ questions.