, 2019

The answer is $64$. Assume that $n \geq 65$. Then there are at least $\binom{65}{2} = 2080$ unordered pairs of subsets $A_i, A_j$. By conditions, to each unordered pair $A_i, A_j$ we can correspond an element $c(i, j) \in S$ such that $c(i, j) \notin A_i \cup A_j$. Since $2080 > 2019$, for some $A_k, A_l$ and $A_p, A_q$ we have $c(k, l) = c(p, q)$. Finally, since at least three of the indices $k, l, p, q$ are distinct, the union of some three subsets is not equal to $S$, a contradiction.

Now let us give an example for $n = 64$. Let $S = \{a_1, a_2, \dots, a_{2019}\}$. There are $\binom{64}{2} = 2016$ unordered pairs of distinct indices $i, j$, $1 \leq i, j \leq 64$. Let us fix any one-to-one correspondence between the set of $2016$ unordered pairs and the set $\{1, 2, \dots, 2016\}$: $A_i, A_j \leftrightarrow m(i, j) \in \{1, 2, \dots, 2016\}$. We start with $A_1 = A_2 = \dots = A_{64} = S$ and after that for each $m(i, j)$ remove $a_{m(i,j)}$ from both $A_i$ and $A_j$. Since each element of $\{a_1, a_2, \dots, a_{2016}\}$ is removed exactly from two subsets, the obtained collection $A_1, A_2, \dots, A_{64}$ satisfies the conditions. Done.