Belarus2022

Note that the sum of all numbers is $0$, so the required sum can be presented as the sum of the following two sums: $$A_1=|a_1|+|a_1+a_2|+\cdots +|a_1+a_2+\cdots +a_{1011}|$$ and $$A_2=|a_{2022}|+|a_{2022}+a_{2021}|+\cdots +|a_{2022}+\cdots +a_{1013}|.$$ Let's bound each term of each sum separately. If $a_{i_1},a_{i_2},\dots ,a_{i_k}$ are $k<1012$ pairwise distinct elements of the sequence given in the problem statement, then $$|a_{i_1}+a_{i_2}+\cdots +a_{i_k}|\le |1011+1010+\cdots +1012-k|$$ Moreover, this estimate is reached when $a_i=1012-i$ or $a_i=-1012+i$. Hence $$A_1\le 1011+(1011+1010)+\cdots +(1+2+\cdots +1011)$$ $$A_2\le 1011+(1011+1010)+\cdots +(1+2+\cdots +1011)$$ So, the maximal possible value of the original sum is equal to $$\begin{gathered} 2\cdot 1011+2\cdot (1011+1010)+\cdots +2\cdot (1011+\cdots +2)+(1011+\cdots +1)= \\ =2\cdot (1^2+2^2+\cdots +1011^2)-(1+2+\cdots +1011)= \\ 2\cdot \frac{1011\cdot 1012\cdot 2023}{6}-\frac{1011\cdot 1012\cdot 3}{6}=\frac{1011\cdot 1012\cdot 4043}{6}. \end{gathered}$$

This sum is reachable if the numbers are arranged in ascending or descending order.