China-TST-B

Proof. Such a sequence does not exist. We prove this by contradiction. Suppose there exists such a sequence. Take a positive integer $N$ satisfying
$$\frac{1}{N+1} + \frac{1}{N+2} + \dots + \frac{1}{N^2} > 2024.$$
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
$$2 \sum_{k=N+1}^{N^2} \left\lfloor \frac{N^2}{k} \right\rfloor \ge 2N^2 \sum_{k=N+1}^{N^2} \frac{1}{k} - 2(N^2 - N) > 2N^2 \cdot 2024 - 2N^2 + 2N > 4046N^2 + 2.$$
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. $\square$