On $2020\times 2021$ squares, cover the whole area with several L-Trominos and S-Tetrominos (=Z-Tetrominos) along the square so that they do not overlap. The tiles can be rotated or flipped over. Find the minimum possible number of L-Trominos to be used.
1. **Define the problem and variables:**
   - We have a $2020 \times 2021$ grid.
   - We need to cover the grid using L-Trominos and S-Tetrominos (Z-Tetrominos) without overlapping.
   - Let \( x \) be the number of S-Tetrominos and \( y \) be the number of L-Trominos.

2. **Calculate the total area:**
   - The total area of the grid is \( 2020 \times 2021 \).

3. **Calculate the area covered by the tiles:**
   - Each S-Tetromino covers 4 squares.
   - Each L-Tromino covers 4 squares.
   - Therefore, the total area covered by \( x \) S-Tetrominos and \( y \) L-Trominos is \( 4x + 4y \).

4. **Set up the equation for the total area:**
   \[
   4x + 4y = 2020 \times 2021
   \]
   Simplifying, we get:
   \[
   x + y = \frac{2020 \times 2021}{4}
   \]

5. **Calculate the number of squares in even columns:**
   - There are 1010 even columns (since \( 2021 \) is odd, half of the columns are even).
   - Each even column has 2020 squares.
   - Therefore, the total number of squares in even columns is \( 2020 \times 1010 \).

6. **Set up the equation for the even columns:**
   - Each S-Tetromino covers exactly 2 squares in even columns.
   - Each L-Tromino covers at least 1 square in even columns.
   - Therefore, the total number of squares covered in even columns is \( 2x + y \).

7. **Set up the inequality for the even columns:**
   \[
   2020 \times 1010 \geq 2x + y
   \]

8. **Combine the equations:**
   - From the total area equation: \( x + y = \frac{2020 \times 2021}{4} \).
   - From the even columns inequality: \( 2020 \times 1010 \geq 2x + y \).

9. **Solve for \( y \):**
   - Substitute \( y = \frac{2020 \times 2021}{4} - x \) into the inequality:
     \[
     2020 \times 1010 \geq 2x + \left( \frac{2020 \times 2021}{4} - x \right)
     \]
   - Simplify the inequality:
     \[
     2020 \times 1010 \geq \frac{2020 \times 2021}{4} + x
     \]
     \[
     2020 \times 1010 - \frac{2020 \times 2021}{4} \geq x
     \]
   - Calculate the right-hand side:
     \[
     2020 \times 1010 - \frac{2020 \times 2021}{4} = 2020 \times 1010 - 505 \times 2021
     \]
     \[
     = 2020 \times 1010 - 505 \times (2020 + 1)
     \]
     \[
     = 2020 \times 1010 - 505 \times 2020 - 505
     \]
     \[
     = 2020 \times (1010 - 505) - 505
     \]
     \[
     = 2020 \times 505 - 505
     \]
     \[
     = 505 \times (2020 - 1)
     \]
     \[
     = 505 \times 2019
     \]
   - Therefore:
     \[
     x \leq 505 \times 2019
     \]

10. **Find the minimum number of L-Trominos:**
    - From \( x + y = \frac{2020 \times 2021}{4} \):
      \[
      y = \frac{2020 \times 2021}{4} - x
      \]
    - To minimize \( y \), maximize \( x \):
      \[
      x = 505 \times 2019
      \]
    - Substitute \( x \) back into the equation:
      \[
      y = \frac{2020 \times 2021}{4} - 505 \times 2019
      \]
      \[
      y = 505 \times 2021 - 505 \times 2019
      \]
      \[
      y = 505 \times (2021 - 2019)
      \]
      \[
      y = 505 \times 2
      \]
      \[
      y = 1010
      \]

The final answer is \( \boxed{1010} \).