Look at an arbitrary olympiad, let that be $A_1$, where the prizes went to some $5$ students. Each of the remaining $21$ olympiads had to have someone among those $5$ receiving a prize. By pigeonhole principle there exists a student who in addition to $A_1$ also got a prize at at least $5$ olympiads. Let that student be $a$ and those olympiads be $A_2, \dots, A_6$.

Let now $B$ be an arbitrary olympiad that is different from $A_1, \dots, A_6$. As each one of the olympiads $A_1, \dots, A_6$ has one prize-winning student in common with $B$ and exactly $5$ students get prizes at $B$, applying pigeonhole principle again shows that one of those five had to get a prize at at least two of $A_1, \dots, A_6$. Since these two have student $a$ in common and according to initial conditions that student is the only one, this means that $a$ also got a prize at olympiad $B$. But since we picked $B$ arbitrarily, $a$ must have got a prize at every olympiad.