Minimal amount of points, that was possible to earn equals to $0$, maximum – to $56$. Consider such segments in points: $[0; 1]$, $[2; 3]$, $[4; 5]$, ..., $[54; 55]$ and $56$ points. If at least 3 students are in at least one of these segments, the statement is proved. If not, then for each of these segments there are not more than two students. There are $29$ segments, so not more than $58$ might have taken part in the Olympiad. This contradiction ends the proof of the first part.

For the second part. If exactly $2$ participants got each of points $0$, $2$, $4$, ..., $56$. Then there are no three, difference of whose results is not more than $1$ point. And in total there are exactly $58$ participants.