The number $2019$ is written on a blackboard.  Every minute, if the number $a$ is written on the board, Evan erases it and replaces it with a number chosen from the set
$$ \left\{ 0, 1, 2, \ldots, \left\lceil 2.01 a \right\rceil \right\} $$
uniformly at random. Is there an integer $N$ such that the board reads $0$ after $N$ steps with at least $99\%$ probability?
1. **Define the probability function**: Let \( p_a^{(N)} \) be the probability that the number \( 0 \) is written \( N \) steps after \( a \) is written. Initially, we have \( p_a^{(0)} = \delta_{a0} \), where \( \delta_{ij} = 1 \) if \( i = j \) and \( 0 \) otherwise.

2. **Recursive relation**: We have the recursive relation:
   \[
   p_a^{(N)} = \frac{1}{\lceil 2.01a \rceil + 1} \left[ p_0^{(N-1)} + p_1^{(N-1)} + \cdots + p_{\lceil 2.01a \rceil}^{(N-1)} \right].
   \]

3. **Initial claim**: The claim is that \( p_a^{(N)} \ge \frac{N}{\lceil 2.01a \rceil + N} \) for \( a \ge 1 \).

4. **Base case**: For \( N = 0 \), \( p_a^{(0)} = 0 \) for \( a \neq 0 \), which satisfies the inequality since \( \frac{0}{\lceil 2.01a \rceil + 0} = 0 \).

5. **Inductive step**: Assume the claim is true for \( N-1 \). We need to show it holds for \( N \). Let \( k = \lceil 2.01a \rceil \). Then:
   \[
   p_a^{(N)} \ge \frac{1}{k+1} \left[ 1 + \frac{k(N-1)}{k+N-1} \right].
   \]

6. **Simplify the expression**:
   \[
   \frac{1}{k+1} \left[ 1 + \frac{k(N-1)}{k+N-1} \right] = \frac{1}{k+1} \cdot \frac{N(k+1) - 1}{k+N-1} = \frac{N}{k+N-1} - \frac{1}{(k+1)(k+N-1)}.
   \]

7. **Compare with the desired bound**:
   \[
   \frac{N}{k+N-1} - \frac{1}{(k+1)(k+N-1)} \ge \frac{N}{k+N} \iff \frac{N}{(k+N-1)(k+N)} \ge \frac{1}{(k+1)(k+N-1)}.
   \]

8. **Simplify the inequality**:
   \[
   \frac{N}{(k+N-1)(k+N)} \ge \frac{1}{(k+1)(k+N-1)} \iff N(k+1) \ge k+N \iff (N-1)k \ge 0.
   \]

9. **Conclusion of the inductive step**: Since \( (N-1)k \ge 0 \) is true for \( N \ge 1 \), the inductive step holds. Thus, \( p_a^{(N)} \ge \frac{N}{\lceil 2.01a \rceil + N} \).

10. **Probability bound**: For \( a = 2019 \), we have:
    \[
    p_{2019}^{(N)} \ge \frac{N}{N + 4059}.
    \]

11. **Determine \( N \) for 99% probability**: We need \( \frac{N}{N + 4059} \ge 0.99 \). Solving for \( N \):
    \[
    \frac{N}{N + 4059} \ge 0.99 \implies N \ge 0.99(N + 4059) \implies N \ge 0.99N + 4018.41 \implies 0.01N \ge 4018.41 \implies N \ge 401841.
    \]

Thus, there exists an integer \( N \) such that the board reads \( 0 \) after \( N \) steps with at least 99% probability.

The final answer is \( \boxed{ N \ge 401841 } \).