Answer is 64 and here is why. WLOG S={1,2,3,...,2019} For every two sets there must exist an elemnt of S that doesn't belong to either group otherwise we have the union of these two groups equals S. The element that doesn't belong to the groups Ai and Aj will be marked as Mi,j (if there is more than 1 element satisfying it, we choose one of them, it doesn't matter which one) If we have two equal numbers in the series M then we have the union of some 3 sets not containing that number so that union isn't equal S which is a contradiction. Therefore all numbers Mi,j are different and they are all elements of S. There are n(n-1)/2 numbers in the series M then we have n(n-1)/2≤2019 implying that n<65 For n=64 we choose for each two sets Ai and Aj one element that belongs to all sets other then Ai and Aj and we are done. Q.E.D

I got this problem from a friend of mine. Here's my solution (which is somewhat similar to #2): The answer is $64$. We are going to prove that this is the maximum possible value of $n$. Let $S=\{a_1,a_2,\dots,a_{2019}\}$. We observe a couple things: $(1)$ Each $a_i$ $(1\le{i}\le{2019})$ is absent in at most two sets. $(2)$ For each pair $A_j,A_k$ $(j\neq{k})$, there is at least one $a_i$ such that $a_i\notin{A_j,A_k}$. Also note that these $a_i$-s are unique to each pair of subsets $A_j,A_k$. Now we are going to make a $2019 \times n$ matrix, and write $1$ in the $i$-th row of the $j$-th column if $a_i\in{A_j}$ and write zero otherwise. By $(1)$ we can say that the sum of the rows is at least $2019(n-2)$. Using $(2)$ on all $\binom{n}{2}$ pairs of subsets we get at least $\binom{n}{2}$ pairs of zeroes. So, the sum of the columns is at most $2019n-2\binom{n}{2}$. Now we have the inequality $$2019(n-2) \le 2019n-2\binom{n}{2},$$which simplifies to $$n(n-1)\le{4038},$$which is true for $n\le{64}$, and we have our maximum value.

We see that for any two subsets ,the elements that they both lack are unique ,because otherwise we would find three subsets whose union is not $S$. For any two subsets ,we get atleast one element of $S$. Thus, $\binom{n}{2} \le 2019$,which just gives $n \le 64$.