Just choose them. So starting from any two points we choose, if there are $k$ chosen points at the moment say our set is $S$, then there are $\dbinom{k}{2}$ distinct lines formed by any two points in our set $S$, by the condition, there will be at most $\dbinom{k}{2}$ points that we havent chose, will lie on these lines. So if $k\le 62$, then there are at least $2016-k-\dbinom{k}{2}\ge 1$ points left that we can put in $S$. So we can find a set $S$ of at least $63$ points with the condition.

Similar to Italy tst problem that appears in Pranav Sriram's Olympiad combinatorics. PS: that problem is a generalization of the original one.