Belarus competition problems

Note that the sum of all numbers is $0$, so the required sum can be presented as the sum of the following two sums:
$$A_1 = |a_1| + |a_1 + a_2| + \dots + |a_1 + a_2 + \dots + a_{1011}|$$
and
$$A_2 = |a_{2022}| + |a_{2022} + a_{2021}| + \dots + |a_{2022} + \dots + a_{1013}|.$$
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
$$|a_{i_1} + a_{i_2} + \dots + a_{i_k}| \le |1011 + 1010 + \dots + 1012 - k|$$
Moreover, this estimate is reached when $a_i = 1012 - i$ or $a_i = -1012 + i$. Hence
$$A_1 \le 1011 + (1011 + 1010) + \dots + (1 + 2 + \dots + 1011)$$
$$A_2 \le 1011 + (1011 + 1010) + \dots + (1 + 2 + \dots + 1011)$$
So, the maximal possible value of the original sum is equal to
$$2 \cdot 1011 + 2 \cdot (1011 + 1010) + \dots + 2 \cdot (1011 + \dots + 2) + (1011 + \dots + 1) = \\ = 2 \cdot (1^2 + 2^2 + \dots + 1011^2) - (1+2+\dots+1011) = \\ 2 \cdot \frac{1011 \cdot 1012 \cdot 2023}{6} - \frac{1011 \cdot 1012 \cdot 3}{6} = \frac{1011 \cdot 1012 \cdot 4043}{6}.$$

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