Obviously, we can always split a group with more than 1 person into two groups following the same discussion pattern, so it's okay if we come up with less groups than requested. Induct on the statement: given no more than $\leq n^2/2$ movie fans, they can be divided into no more than $n$ groups. $n = 1$: trivial, because there are no fans. Now, suppose we have $\leq n^2/2$ fans to be split into $\leq n$ groups. Pick a minimal set of movies $M$ such that each fan has watched at least one of these movies. If there are no more than $n$ movies in $M$, we can make one group per movie and we're done. If there are more than $n$ movies, then by the minimalness of $M$, for each movie in $M$ we must have some fan that has watched no movie of $M$ other than that one. These fans, numbering more than $n$, can form a group discussing these movies; we can then use the induction hypothesis on the remaining fans, numbering $\leq (n-1)^2/2$.

I have a solution that improves the bound to $<\frac {(n+1)(n+2)}{2}$ movie fans for $n$ groups. Call groups where all the people watched the same movie "same group" and where each has a unique movie "unique group". We use induction. For $n=1$ there are either $1$ or $2$ movie fans so obviously we can put them in $1$ "same group" or "unique group". Now suppose it is true that if there are $<\frac {n(n+1)}{2}$ movie fans then we can put them in $n-1$ groups. Suppose that each movie was watched by at most $n$ people, then we can definitely form $n$ "unique groups". Number the movies and mentally create $n$ empty groups, first put people of movie $1$ into different groups. Next put movie $2$ in different groups while avoiding some people who liked movie $1$ and $2$. Then put people who watched movie $3$ into different groups while avoiding some people who watched $3$ plus movie $1$ and/or movie $2$ and continue in this manner. Since each movie was watched by at most $n$ people it is always possible to do this, and for each person the lowest numbered movie he/she watched will be the movie that no one else in his/her group watched. Otherwise, there is a movie watched by at least $n+1$ people, so make a "same group" using them. Now there are $n-1$ groups remaining and $<\frac {n(n+1)}{2}$ movie fans, so we apply the inductive hypothesis. Strangely, the official solution uses pretty much the same idea as me but they got a lower bound of $\frac {n(n+1)}{2}$ so I'm wondering if I made a made a mistake somewhere.

I'll prove by induction on $x$ that $2x^2$ participants can be splitted into $2x$ groups.For $x=1$ just put one guy in one group and the other guy in the other.Suppose it hold for $x-1$.Set up incidence matrix where the rows are movies and and columns are the people.Let the movies be $c_1,c_2,...c_k$ and the people $1,2...2x^2$ and $a_{ij}=1$ if the movie $c_i$ was watched by the guy $j$ and zero otherwise.It suffices to show that we can split $4x-2$ guys in $2$ groups.First suppose there exist a row with more than than one $\{1\}$.Say $c_j$ and let it have $p$ zeroes and $q$ ones.If $q>4x-2$ just split the ones into two groups adding up to $4x-2$ and we finish.If $q\leq 4x-2$ keep adding ones into one group till one remains than just throw in a bunch remaining zeroes till the $4x+2$ condition is fulfilled there doesn't exist row $c_j$ than cay in rows $c_i,c_j$ guys $v,w$ have ones now just form group with $2x-1$ guys from $c_i$ and $2x-1$ people from $c_j$ and so we're done.$\blacksquare$