Iranian Mathematical Olympiad

The answer is $16$. For the example consider the following family of $5$-element subsets.
$$\begin{aligned} \{1, 2, 3, 4, 5\} & & \{2, 6, 10, 14, 18\} \\ \{2, 7, 11, 15, 19\} & & \{2, 8, 12, 16, 20\} \\ \{1, 6, 7, 8, 9\} & & \{3, 8, 13, 15, 18\} \\ \{3, 9, 12, 14, 19\} & & \{3, 6, 11, 17, 20\} \\ \{1, 10, 11, 12, 13\} & & \{4, 7, 12, 17, 18\} \\ \{4, 6, 13, 16, 19\} & & \{4, 9, 10, 15, 20\} \\ \{1, 14, 15, 16, 17\} & & \{5, 9, 11, 16, 18\} \\ \{5, 8, 10, 17, 19\} & & \{5, 7, 13, 14, 20\} \end{aligned}$$
Next, we prove that there is no such family with $17$ subsets. Without loss of generality we can assume that $A_1 = \{1, 2, 3, 4, 5\}$ is among the subsets. By pigeonhole principle there are at least four subsets such that intersection of each of them with $A_1$ is the same. Without loss of generality call these four subsets $A_2, A_3, A_4$ and $A_5$, and assume that for all integer numbers $2 \le i \le 5$, $A_i \cap A_1 = \{1\}$. So by assumption we also know that $A_i \cap A_j = \{1\}$ for all integer numbers $2 \le i < j \le 5$. Hence $A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5$ has $21$ elements which is a contradiction. ■