One year, $n$ (three or more) candidates came out in the election to select the mayor of the island of knights and knaves. In one TV debate, all the candidates stated in turn: For each $k = 1, 2, 3, \ldots, n$, the $k$-th candidate said: Among the candidates here, except me, there are just $k$ more knaves than knights.
Assume that the knaves speak falsely and the knights speak truly. How many candidates are there?
There is at most one knight candidate. If there were at least two distinct ones, each knight would see the same number of knights and knaves other than themselves, hence the same difference between knights and knaves, but the statements of each candidate are mutually exclusive, so at most one can be true, which is a contradiction.
If all the candidates are knaves, candidate $n-1$'s statement is true, which contradicts their statement being false. Thus there is exactly one knight candidate, and it has to be candidate $n-1$ (seeing as the other $n-1$ candidates are knaves).
Yet if $n>3$, candidate $n-3$ would see $n-2$ knaves and one knight besides themselves, hence their statement would be true, which is a contradiction. Thus $n=3$, with candidate 2 being a knight and the other two being knaves.