Mongolian Mathematical Olympiad

Let us assume that the union of the sets consists of integers from $1$ to $n$. Let $S_i$ denote the number of sets that $i$ belongs to. Then the total number of elements of $n$ sets is $5n = S_1 + S_2 + \dots + S_n$ ($\bullet$).

Let us assume $S_1 > 5$ and $1 \in A_1, 1 \in A_2, 1 \in A_3, 1 \in A_4, 1 \in A_5, 1 \in A_6$. If $1 \in A_i, i = 1, \overline{n}$, then the remaining $4n$ elements have to be different. Because $|A_i \cap A_j| = 1, i \neq j$ holds. That means, we have $4n+1$ different elements. It contradicts to $|A_1 \cup A_2 \cup \dots \cup A_n| = n$. Thus we can assume that $1 \notin A_7$.

Since $|A_i \cap A_j| = 1$ for $1 \le i, j \le 6$, the intersections of $A_7$ with $A_1, A_2, \dots, A_6$ are all different. Hence $|A_7| \ge 6$. But it contradicts to $|A_7| = 5$. This leads to $S_1 \le 5$. Analogously, $S_k \le 5$ for $k = \overline{1, n}$. Considering $(*)$, $S_k = 5$ holds for $i = \overline{1, n}$. Hence the number of sets is $n = 4 \cdot 5 + 1 = 21$. The construction is:

$$\begin{align*} A_1 &= \{1, 2, 3, 4, 5\} & A_2 &= \{1, 6, 7, 8, 9\} & A_3 &= \{1, 10, 11, 12, 13\} \\ A_4 &= \{1, 14, 15, 16, 17\} & A_5 &= \{1, 18, 19, 20, 21\} & A_6 &= \{2, 6, 10, 14, 18\} \\ A_7 &= \{2, 7, 11, 15, 19\} & A_8 &= \{2, 8, 12, 16, 20\} & A_9 &= \{2, 9, 13, 17, 21\} \\ A_{10} &= \{3, 6, 7, 8, 9\} & A_{11} &= \{3, 10, 11, 12, 13\} & A_{12} &= \{3, 14, 15, 16, 17\} \\ A_{13} &= \{3, 18, 19, 20, 21\} & A_{14} &= \{4, 6, 10, 14, 18\} & A_{15} &= \{4, 7, 11, 15, 19\} \\ A_{16} &= \{4, 8, 12, 16, 20\} & A_{17} &= \{4, 9, 13, 17, 21\} & A_{18} &= \{5, 6, 7, 8, 9\} \\ A_{19} &= \{5, 10, 11, 12, 13\} & A_{20} &= \{5, 14, 15, 16, 17\} & A_{21} &= \{5, 18, 19, 20, 21\}. \end{align*}$$