62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour

Denote by $s_i=a_1+a_2+\cdots +a_i$, $t_i=1+2+\cdots +i$. 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 $s_i\le 2022+2021+\cdots +(2023-i)=2023i-1011$. 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 $s_i-s_j=a_{j+1}+a_{j+2}+\cdots +a_i\le 2022(i-j)$, 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 $i=1,\dots ,1011$ we have $s_i=2+4+\cdots +2i=2t_i$.