There are 27 people. At least half of them are truth-tellers, others are tricksters. Tricksters can answer yes/no regardless of your question. But we should ask them such questions that truth-tellers can answer as well. I would also assume that tricksters also are trying to hide the fact they are tricksters. So if you ask 'are a you a truth-teller' any of them would answer 'yes'.
All of the people know who all the others are (trickster or truth-teller).
We can ask any of them any yes/no questions. You can assume the questions you are asking are not heard by other people.
We need to identify at least one of the truth-tellers using minimum questions.
I have a strategy for 25 questions but there is a better strategy that I am looking for. Ideally I would like to find an optimal amount of questions with a proof that it is not possible to solve the puzzle with less questions. But for now I am looking at least for strategy with less than 25 questions.
My strategy for 25 questions:
Prove by induction that if we have 2N+1 people with at most N tricksters, we have a strategy with 2N-1 questions.
Base N=1. We have 3 people with at most 1 trickster. We ask person #1 'Is person #2 a truth-teller?'. If answer is 'yes' - person #2 is a truth-teller, if 'no' - person #3 is a truth-teller. We solved it with 2N-1=1 questions. Let's see why this strategy works. If #1 is a trickster and as we know there are at most 1 trickster, #2 and #3 are truth-tellers so our strategy works. If #1 is truth-teller and said 'yes' - #2 is a truth-teller, our strategy works! If #1 is truth-teller and said 'no' - #2 is a trickster, so #3 is a truth-teller, our strategy works!
Now, assuming we proved the statement for N, consider N+1
We have 2N+3 people with at most N+1 tricksterstricksters. We need to prove that we have a strategy with 2N+1 questions.
We ask person #1 'Is person #(2N+3) a truth-teller?'.
a) If answer is 'no', then definitely either #1 or #(2N+3) is a trickster (or both). So we exclude them and we have 2N+1 people (#2,...,#(2N+2)) with at most N tricksters. We already used one question and by induction we will need 2N-1 questions to identify a truth-teller. So in total we used 1+(2N-1)=2N<2N+1 questions
b) If answer is 'yes', we keep asking other people the same question about #(N+1).
If we got N+1 'yes' answers, then person #(2N+3) truly is a truth-teller, because only N+1 people could potentially lie. We used N+1 answers N+1<2N+1
c) If we got #1, ..., #k replied 'yes' and #(k+1) replied 'no', k<=N, it means, either #k or #(k+1) is a trickster. Exclude both of them from consideration. We have 2N+1 people with at most N tricksters. We already used k+1 questions, but first k-1 are exactly the same we would ask if we have those people (1,2,...,k-2,k-1,k+2,...,2N+3) from the very beginning. So actually only two questions should be thrown away. By induction, we will need 2N-1 questions. Adding those two thrown away, we will get 2+(2N-1)=2N+1
In our case N=13, 27=2N+1 and we need 2N-1=25 questions!
What is the optimal strategy?
UPD: I found a related puzzle, but there is required to find all truth-tellers. In our puzzle we need to find only one. Knights and jokers