On a table near the sea, there are $N$ glass boxes where $N<2021$, each containing exactly $2021$ balls. Sowdha and Rafi play a game by taking turns on the boxes where Sowdha takes the first turn. In each turn, a player selects a non-empty box and throws out some of the balls from it into the sea. If a player wants, he can throw out all of the balls in the selected box. The player who throws out the last ball wins. Let $S$ be the sum of all values of $N$ for which Sowdha has a winning strategy and let $R$ be the sum of all values of $N$ for which Rafi has a winning strategy. What is the value of $\frac{R-S}{10}$?
1. **Determine the winning strategy for each player:**
   - If $N$ is even, Rafi wins by copying Sowdha's moves. This is because after each of Sowdha's moves, Rafi can always make a move that leaves an even number of boxes, maintaining the symmetry.
   - If $N$ is odd, Sowdha wins by removing all balls from one pile. This leaves an even number of boxes for Rafi, effectively making Rafi the first player in a game with an even number of boxes, which we already know is a losing position for the first player.

2. **Sum the values of $N$ for which each player has a winning strategy:**
   - For Rafi to win, $N$ must be even. The even numbers less than 2021 are $2, 4, 6, \ldots, 2020$. This is an arithmetic series with the first term $a = 2$, the common difference $d = 2$, and the last term $l = 2020$.
     - The number of terms in this series is given by:
       \[
       n = \frac{2020 - 2}{2} + 1 = 1010
       \]
     - The sum of the first $n$ terms of an arithmetic series is given by:
       \[
       S_n = \frac{n}{2} (a + l) = \frac{1010}{2} (2 + 2020) = 505 \times 2022 = 1021110
       \]
     - Therefore, $R = 1021110$.

   - For Sowdha to win, $N$ must be odd. The odd numbers less than 2021 are $1, 3, 5, \ldots, 2019$. This is an arithmetic series with the first term $a = 1$, the common difference $d = 2$, and the last term $l = 2019$.
     - The number of terms in this series is given by:
       \[
       n = \frac{2019 - 1}{2} + 1 = 1010
       \]
     - The sum of the first $n$ terms of an arithmetic series is given by:
       \[
       S_n = \frac{n}{2} (a + l) = \frac{1010}{2} (1 + 2019) = 505 \times 2020 = 1020100
       \]
     - Therefore, $S = 1020100$.

3. **Calculate $\frac{R - S}{10}$:**
   \[
   R - S = 1021110 - 1020100 = 1010
   \]
   \[
   \frac{R - S}{10} = \frac{1010}{10} = 101
   \]

The final answer is $\boxed{101}$.