Ann and Drew have purchased a mysterious slot machine; each time it is spun, it chooses a random positive integer such that $k$ is chosen with probability $2^{-k}$ for every positive integer $k$, and then it outputs $k$ tokens. Let $N$ be a fixed integer. Ann and Drew alternate turns spinning the machine, with Ann going first. Ann wins if she receives at least $N$ total tokens from the slot machine before Drew receives at least $M=2^{2018}$ total tokens, and Drew wins if he receives $M$ tokens before Ann receives $N$ tokens. If each person has the same probability of winning, compute the remainder when $N$ is divided by $2018$.

[i]Proposed by Brandon Wang[/i]
1. **Problem Translation**: We need to determine the value of \( N \) such that the probability of Ann reaching \( N \) tokens before Drew reaches \( M = 2^{2018} \) tokens is \( \frac{1}{2} \). This can be visualized as a particle moving on a grid, starting at \((0, 0)\), moving up or right with equal probability, and we need to find \( N \) such that the particle hits the line \( y = N \) before \( x = M \) with probability \( \frac{1}{2} \).

2. **Grid Representation**: Consider an \((M+1) \times (N+1)\) grid. The particle starts at \((0, 0)\) and moves up or right with equal probability. We need to find \( N \) such that the probability of hitting the line \( y = N \) before the line \( x = M \) is \( \frac{1}{2} \).

3. **Probability Calculation**: Let \( m \) be the number of ways to reach the line \( x = M \) before the line \( y = N \), and \( n \) be the number of ways to reach the line \( y = N \) before the line \( x = M \). We need \( m = n \).

4. **Counting Paths**:
   - The number of ways to reach the line \( y = N \) before \( x = M \) is given by:
     \[
     n = \sum_{k=0}^{M-1} \binom{N+k-1}{N-1} = \binom{N+M-1}{N}
     \]
   - The number of ways to reach the line \( x = M \) before \( y = N \) is given by:
     \[
     m = \sum_{k=0}^{N-2} \binom{M+k}{M} = \binom{N+M-1}{M+1}
     \]

5. **Equating the Paths**:
   \[
   \binom{N+M-1}{M+1} = \binom{N+M-1}{N}
   \]
   Using the property of binomial coefficients, we have:
   \[
   \binom{N+M-1}{M+1} = \binom{N+M-1}{N}
   \]
   This implies:
   \[
   M + 1 = N
   \]
   Substituting \( M = 2^{2018} \):
   \[
   N = 2^{2018} + 1
   \]

6. **Finding the Remainder**: We need to find the remainder when \( N \) is divided by \( 2018 \):
   \[
   N = 2^{2018} + 1
   \]
   Using Euler's theorem, since \( \phi(2018) = 1008 \):
   \[
   2^{2018} \equiv 2^{2 \cdot 1009} \equiv (2^{1008})^2 \equiv 1^2 \equiv 1 \pmod{2018}
   \]
   Therefore:
   \[
   2^{2018} + 1 \equiv 1 + 1 \equiv 2 \pmod{2018}
   \]

The final answer is \( \boxed{2} \).