Order the lectures in decreasing order of size and arrange one lecture on each day. Consider a lecture $L$, with say $m$ people, that occurs after day $211$. I will show that it can have been moved to an earlier day. Repeating this argument will give that all lectures can be moved to one of the first $211$ days. Suppose not, then there must be a lecture on each previous day that clashes with it, so at least $211$ of them. By PHP, at least $\left \lceil \frac{211}{m} \right \rceil$ of them must overlap with $L$ at the same element, say $M$. Then, since no two lectures can share more than one person, apart from $M$, all of these lectures must be disjoint. Since the lectures are arranged so all later ones have at most as much as previous ones, every other lecture in consideration also has at least $m$ people in it. However, observe that $\left \lceil \frac{211}{m} \right \rceil\cdot (m-1) + m > 200$ for $m \geqslant 10$ (easy to verify). This is a contradiction since there were only $200$ students. Therefore, we can move lecture $L$ earlier and repeat this to finish the problem. $\blacksquare$