Ukrainian National Mathematical Olympiad, Third Round, Second Tour

Denote by $s_i = a_1 + a_2 + \dots + a_i$, $t_i = 1 + 2 + \dots + i$. We will show that there exists at most one $i \ge 1011$, for which $s_i \ne t_i$ is divisible by $t_i$. Indeed, note that $s_i \le 2022 + 2021 + \dots + (2023 - i) = 2023i - 1011$. Also, for $i \ge 1011$ we have $2023i - 1011 < 3t_i$, as $4t_i = 2i(i+1) \ge 2024i$, so if for $i \ge 1011$ $t_i$ is an integer larger than $1$, then $s_i = 2t_i$.

Suppose that there exist two such $i > j \ge 1011$, that $s_i = 2t_i, s_j = 2t_j$, then $s_i - s_j = 2(t_i - t_j) = i^2 + j^2 - i - j = (i-j)(i+j+1) \ge 2023(i-j)$, but $s_i - s_j = a_{j+1} + a_{j+2} + \dots + a_i \le 2022(i-j)$, contradiction. So, there can't be more than $1011$.

Let it be $i$. It's enough to note that for a permutation $(2, 4, \dots, 2022, 1, 3, \dots, 2021)$ the number of such $i$ is precisely $1011$, as for $i = 1, \ldots, 1011$ we have $s_i = 2+4+\dots+2i = 2t_i$.