Olimpiada Matemática del Cono Sur

We will first show how to construct $n-2$ subsets of $\{1, 2, \dots, n\}$ satisfying the required conditions. Such subsets will be called nice.
For $n=4$, we can take $A_1 = \{1\}$ and $A_2 = \{2, 3\}$.
For $n=5$, we can take $A_1 = \{1\}$, $A_2 = \{2, 3\}$ and $A_3 = \{2, 4, 5\}$.
We show now that if there exist $n-2$ nice subsets for $n$, then there exist $n$ nice subsets for $n+2$. Let $A_1, A_2, \dots, A_{n-2}$ be nice subsets for $n$. Consider:
$$\bullet B_1 = \{n+2\},$$
$$\bullet B_{i+1} = A_i \cup \{n+1\}, \text{ for every } 1 \le i \le n-2,$$
$$\bullet B_n = \{1, 2, \dots, n\}.$$
It is easy to check that $B_1, B_2, \dots, B_n$ are nice subsets for $n+2$.
Now, it remains to be seen that it is not possible to construct $n-1$ nice subsets for $n$.
Assume, by contradiction, that $A_1, \dots, A_{n-1}$ are nice subsets for $n$. If $A_1 = \{x\}$, then $A_{n-1} = \{1, \dots, n\} \setminus \{x\}$. Since $A_2$ has two elements that are different from $x$, it follows that they are elements of $A_{n-1}$ and so, $A_2 \subset A_{n-1}$, which is a contradiction.