Saudi Arabia booklet, 2024

Let $d$ be the number of distinct values appear on the table, let them be
$$a_1 < a_2 < \dots < a_d.$$
Assuming by contradiction that no value appears at least $n$ times, then each value appears at most $n-1$ times. So we can count the amount of numbers on the board to get
$$4n - 3 \le d(n - 1) \implies d \ge \frac{4n - 3}{n - 1} \implies d \ge 5.$$
Consider the 5 arbitrary distinct values among them, denoted by $a < b < c < d < e$. According to the assumption, $(a, b, c, d)$ and $(a, b, c, e)$ both form arithmetic progressions. Then, we have $b - a = d - c = e - c$, implies that $d = e$, which is clearly a contradiction. $\square$