China-TST-2023B

Proof. Such a sequence does not exist. We prove this by contradiction. Suppose there exists such a sequence. Take a positive integer $N$ satisfying $$\frac{1}{N+1}+\frac{1}{N+2}+\cdots +\frac{1}{N^2}>2024.$$ Such an $N$ exists because $$\sum _{k=N+1}^{N^2}\frac{1}{k}\ge {\int }_{N+1}^{N^2+1}\frac{1}{k}=\ln \frac{N^2+1}{N+1}>\ln (N-1),$$ which can be arbitrarily large. We prove that at least $4046N^2+2$ elements of $a_1,a_2,\dots$ fall into the interval $S=[-2023N^2,2023N^2]$, which contradicts the assumption that $a_i$'s are all distinct. To show this, it suffices to prove that for $k=N,N+1,\dots ,N^2-1$, (i) At least $\lfloor \frac{N^2}{k+1}\rfloor$ elements of $a_{k^2},\dots ,a_{k^2+k-1}$ fall into $S$. (ii) At least $\lfloor \frac{N^2}{k+1}\rfloor$ elements of $a_{k^2+k},\dots ,a_{k^2+2k}$ fall into $S$. In this way, the total number of elements in $S$ is greater than or equal to $$2\sum _{k=N+1}^{N^2}\lfloor \frac{N^2}{k}\rfloor \ge 2N^2\sum _{k=N+1}^{N^2}\frac{1}{k}-2(N^2-N)>2N^2\cdot 2024-2N^2+2N>4046N^2+2.$$ We will only prove (i), and the proof of (ii) is similar. Note that for $\ell \in \{k^2,k^2+1,\dots ,k^2+k-1\}$, we have $$|a_{\ell +1}-a_{\ell }|\le 2023(k+1).$$ We consider the following three cases. (a) If there exists an $a_{\ell }\ge 2023N^2$ in the sequence $a_{k^2},\dots ,a_{k^2+k-1}$, then () implies that there are at least $\lfloor \frac{2023N^2}{2023(k+1)}\rfloor$ elements in $[0,2023N^2]$ among $a_{\ell },a_{\ell +1},\dots ,a_{k^2+k-1}$. (b) If there exists an $a_{\ell }\le -2023N^2$ in the sequence $a_{k^2},\dots ,a_{k^2+k-1}$, then () implies that there are at least $\lfloor \frac{2023N^2}{2023(k+1)}\rfloor$ elements in $[-2023N^2,0]$ among $a_{k^2},a_{k^2+1},\dots ,a_{\ell }$. (c) If neither (a) nor (b) holds, then all the elements in the sequence $a_{k^2},\dots ,a_{k^2+k-1}$ are within $S$, and there are a total of $k\ge \lfloor \frac{N^2}{k+1}\rfloor$ elements. This completes the proof of (i), and the proof of (ii) can be similarly established. Therefore, a contradiction is reached, and thus such a sequence does not exist. $\square$