TSTST 2024 Problem 3

Let $A = \{a_{1},\ldots ,a_{2024}\}$ be a set of 2024 pairwise distinct real numbers. Assume that there exist positive integers $b_{1},b_{2},\ldots ,b_{2024}$ such that

$$a_{1}b_{1} + a_{2}b_{2} + \cdot \cdot \cdot +a_{2024}b_{2024} = 0.$$

Prove that one can choose $a_{2025},a_{2026},a_{2027},\ldots$ such that $a_{k}\in A$ for all $k\geq 2025$ and, for every positive integer $d$ , there exist infinitely many positive integers $n$ satisfying

$$\sum_{k = 1}^{n}a_{k}k^{d} = 0.$$