, 2023

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

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