SAUDI ARABIAN IMO Booklet 2023

Note that $19$, $23$ are primes and $19\cdot 23=437$. Now denote $S=\{1,2,3,\dots ,20\}$ then take $T=\{437a\mid a\in S\}\subset \{1,2,\dots ,10^4\}$. So $|T|=|S|=20$. Note that $1+2+\cdots +20=210<437$.

Suppose that there exist $k$ numbers in $T$, denote by $437a_1,437a_2,\dots ,437a_k$ such that their sum is a perfect power. Thus $$437(a_1+a_2+\cdots +a_k)=a^n(\ast )$$ for some positive integer $a,n$ and $n\ge 2$. Since $437\mid a^n$ then $437^2\mid a^n$ which implies that $$437\mid a_1+a_2+\cdots +a_k,$$ this is contradiction since the sum of all numbers in $S$ is less than $437$. Thus, the above set satisfies the given condition. $\square$