You have a $3 \times 2021$ chessboard from which one corner square has been removed. You also have a set of $3031$ identical dominoes, each of which can cover two adjacent chessboard squares. Let $m$ be the number of ways in which the chessboard can be covered with the dominoes, without gaps or overlaps.
What is the remainder when $m$ is divided by $19$?
1. **Define the sequences:**
   Let \( t_{2k+1} \) represent the number of ways to cover a \( 3 \times (2k+1) \) chessboard with one corner square removed, and let \( s_{2k} \) represent the number of ways to cover a \( 3 \times 2k \) chessboard with no corner squares removed.

2. **Establish recurrence relations:**
   From the problem, we have the following recurrence relations:
   \[
   t_{2k+1} = s_{2k} + t_{2k-1}
   \]
   \[
   s_{2k} = s_{2k-2} + 2t_{2k-1}
   \]

3. **Initial conditions:**
   From visual inspection, we have:
   \[
   t_1 = 1 \quad \text{and} \quad t_3 = 4
   \]

4. **Derive a simplified recurrence relation:**
   By manipulating the two equations, we can derive:
   \[
   t_{2k+1} = 4t_{2k-1} - t_{2k-3}
   \]
   Define the sequence \( \{u_n\} \) by \( u_n = t_{2n-1} \). Thus, we need to find \( u_{1011} \mod 19 \).

5. **Restate the recurrence relation:**
   The sequence \( \{u_n\} \) follows:
   \[
   u_{n+1} = 4u_n - u_{n-1}
   \]
   with initial conditions:
   \[
   u_1 = 1 \quad \text{and} \quad u_2 = 4
   \]

6. **Calculate the sequence modulo 19:**
   We calculate the first few terms modulo 19:
   \[
   \begin{aligned}
   u_1 & = 1, \\
   u_2 & = 4, \\
   u_3 & = 4u_2 - u_1 = 4 \cdot 4 - 1 = 15, \\
   u_4 & = 4u_3 - u_2 = 4 \cdot 15 - 4 = 60 - 4 = 56 \equiv 18 \mod 19, \\
   u_5 & = 4u_4 - u_3 = 4 \cdot 18 - 15 = 72 - 15 = 57 \equiv 0 \mod 19, \\
   u_6 & = 4u_5 - u_4 = 4 \cdot 0 - 18 = -18 \equiv 1 \mod 19, \\
   u_7 & = 4u_6 - u_5 = 4 \cdot 1 - 0 = 4 \mod 19.
   \end{aligned}
   \]
   We observe that the sequence starts repeating with \( u_6 = 1 \) and \( u_7 = 4 \).

7. **Identify the period of the sequence:**
   The sequence \( \{u_n\} \) repeats every 6 terms:
   \[
   u_1 \equiv 1, \quad u_2 \equiv 4, \quad u_3 \equiv 15, \quad u_4 \equiv 18, \quad u_5 \equiv 0, \quad u_6 \equiv 1 \mod 19.
   \]

8. **Find \( u_{1011} \mod 19 \):**
   Since the sequence repeats every 6 terms, we find the position of \( u_{1011} \) in the cycle:
   \[
   1011 \mod 6 = 5.
   \]
   Therefore, \( u_{1011} \equiv u_5 \mod 19 \).

9. **Conclusion:**
   From the sequence, \( u_5 \equiv 0 \mod 19 \).

The final answer is \( \boxed{0} \).