Romanian Mathematical Olympiad

Notice that $A_n \subset \mathbb{Z}$ for all $n \in \mathbb{N}$, $n \ge 2$.

a) The elements of $A_2$ satisfy the inequalities $x - 2 < 2x \leq x$. By inspection, we obtain $A_2 = \{-1, 0\}$. The elements of $A_3$ satisfy the inequalities $5x - 12 < 6x \leq 5x$. By inspection, we have $A_3 = \{-7, -5, -4, -3, -2, 0\}$, so $A_2 \cup A_3 = \{-7, -5, -4, -3, -2, -1, 0\}$.

b) For $n \ge 4$ and $x \in A_n$ we have $x \le \left(\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right) x$.
From $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} > 1$ we obtain $x \ge 0$.
For $x, n \in \mathbb{Z}$ and $n \ge 2$ we get $\lfloor \frac{x}{n} \rfloor \ge \frac{x - (n - 1)}{n}$. Therefore, if $n \ge 4$ and $x \in A_n$, then
$$x \ge \lfloor \frac{x}{2} \rfloor + \lfloor \frac{x}{3} \rfloor + \lfloor \frac{x}{4} \rfloor \ge \frac{x-1}{2} + \frac{x-2}{3} + \frac{x-3}{4},$$
implying $x \le 23$. Hence the set $A$ is upper bounded.
Because $A \subset \{-5, -4, \dots, 23\}$ and $23 \in A_4$, then $\max A = 23$.