Saudi Arabia booklet 2024

Let $d$ be the number of distinct values appear on the table, let them be $$a_1<a_2<\cdots <a_d.$$ 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 $$4n-3\le d(n-1)⟹d\ge \frac{4n-3}{n-1}⟹d\ge 5.$$ 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. $\square$