, 2022

Reference Solution 1 (By contradiction). Suppose that there exist $2022$ positive integers $x_0 < x_1 < \ldots < x_{2021}$ that violate the problem statement. Then in particular $x_{2021} + 2x_i \le 3x_{i+1}$ for all $i = 0, \ldots, 2020$ which gives
$$x_{2021} - x_i \ge \frac{3}{2}(x_{2021} - x_{i+1}).$$
By a trivial induction we then get
$$x_{2021} - x_i \ge \left(\frac{3}{2}\right)^{2020-i} (x_{2021} - x_{2020}),$$
which for $i = 0$ yields the contradiction since
$$x_{2021} - x_0 \ge \left(\frac{3}{2}\right)^{2020} (x_{2021} - x_{2020}) = \left(\frac{81}{16}\right)^{505} (x_{2021} - x_{2020}) > 5^{505}.$$
Therefore, the proof is complete. $\square$

Reference Solution 2. Denote the maximum element of $A$ by $c$. For $k = 0, \ldots, 2019$, let
$$A_k = \{x \in A : (1 - (2/3)^k)c \le x < (1 - (2/3)^{k+1})c\}.$$
Note that
$$(1 - (2/3)^{2020})c = c - (16/81)^{505}c > c - (1/5)^{505}c \ge c - 1,$$
which shows that the sets $A_0, A_1, \ldots, A_{2019}$ form a partition of $A\setminus\{c\}$. Since $A\setminus\{c\}$ has $2021$ elements, by the pigeonhole principle some set $A_k$ does contain at least two elements of $A\setminus\{c\}$. Denote these two elements $a$ and $b$ and assume $a < b$, so that $a < b < c$. Then
$$c + 2a \ge c + 2(1 - (2/3)^k)c = (3 - 2(2/3)^k)c = 3(1 - (2/3)^{k+1})c > 3b,$$
as desired. $\square$