Let $r_1, r_2, ..., r_{2021}$ be the not necessarily real and not necessarily distinct roots of $x^{2022} + 2021x = 2022$. Let $S_i = r_i^{2021}+2022r_i$ for all $1 \le i \le 2021$. Find $\left|\sum^{2021}_{i=1} S_i \right| = |S_1 +S_2 +...+S_{2021}|$.
1. Given the polynomial equation \( x^{2022} + 2021x = 2022 \), we need to find the roots \( r_1, r_2, \ldots, r_{2022} \).
2. For each root \( r_i \), we have:
   \[
   r_i^{2022} + 2021r_i = 2022
   \]
   Solving for \( r_i^{2021} \), we get:
   \[
   r_i^{2021} = \frac{2022 - 2021r_i}{r_i} = \frac{2022}{r_i} - 2021
   \]
3. Define \( S_i = r_i^{2021} + 2022r_i \). Substituting the expression for \( r_i^{2021} \), we get:
   \[
   S_i = \left( \frac{2022}{r_i} - 2021 \right) + 2022r_i = \frac{2022}{r_i} + 2022r_i - 2021
   \]
4. We need to find the sum \( \left| \sum_{i=1}^{2022} S_i \right| \):
   \[
   \sum_{i=1}^{2022} S_i = \sum_{i=1}^{2022} \left( \frac{2022}{r_i} + 2022r_i - 2021 \right)
   \]
   This can be split into three separate sums:
   \[
   \sum_{i=1}^{2022} S_i = 2022 \sum_{i=1}^{2022} \frac{1}{r_i} + 2022 \sum_{i=1}^{2022} r_i - 2021 \cdot 2022
   \]
5. Notice that the polynomial \( x^{2022} + 2021x - 2022 = 0 \) has roots \( r_1, r_2, \ldots, r_{2022} \). The polynomial \( 2022y^{2022} - 2021y^{2021} - 1 = 0 \) has roots \( \frac{1}{r_1}, \frac{1}{r_2}, \ldots, \frac{1}{r_{2022}} \).
6. By Vieta's formulas, the sum of the roots of the polynomial \( x^{2022} + 2021x - 2022 = 0 \) is zero:
   \[
   \sum_{i=1}^{2022} r_i = 0
   \]
7. Similarly, the sum of the roots of the polynomial \( 2022y^{2022} - 2021y^{2021} - 1 = 0 \) is also zero:
   \[
   \sum_{i=1}^{2022} \frac{1}{r_i} = 0
   \]
8. Substituting these results back into the sum \( \sum_{i=1}^{2022} S_i \):
   \[
   \sum_{i=1}^{2022} S_i = 2022 \cdot 0 + 2022 \cdot 0 - 2021 \cdot 2022 = -2021 \cdot 2022
   \]
9. Taking the absolute value, we get:
   \[
   \left| \sum_{i=1}^{2022} S_i \right| = \left| -2021 \cdot 2022 \right| = 2021 \cdot 2022
   \]

The final answer is \(\boxed{2021 \cdot 2022}\).