Let $n$ be a positive integer. E. Chen and E. Chen play a game on the $n^2$ points of an $n \times n$ lattice grid. They alternately mark points on the grid such that no player marks a point that is on or inside a non-degenerate triangle formed by three marked points. Each point can be marked only once. The game ends when no player can make a move, and the last player to make a move wins. Determine the number of values of $n$ between $1$ and $2013$ (inclusive) for which the first player can guarantee a win, regardless of the moves that the second player makes.

[i]Ray Li[/i]
1. **Understanding the Game Rules**:
   - The game is played on an \( n \times n \) lattice grid.
   - Players alternately mark points on the grid.
   - A point cannot be marked if it is on or inside a non-degenerate triangle formed by three previously marked points.
   - Each point can be marked only once.
   - The game ends when no player can make a move, and the last player to make a move wins.

2. **Strategy for Odd \( n \)**:
   - If \( n \) is odd, the grid has a center point at \((\frac{n+1}{2}, \frac{n+1}{2})\).
   - The first player marks this center point.
   - For any move by the second player, the first player can always choose the reflection of the second player's move across the center point.
   - This strategy ensures that the first player's moves are never inside the convex hull of the marked points, as the reflection symmetry maintains the convex hull's properties.
   - Therefore, the first player can always guarantee a win when \( n \) is odd.

3. **Strategy for Even \( n \)**:
   - If \( n \) is even, there is no single center point, but rather a 2x2 block of central points.
   - The second player can always choose the reflection of the first player's move across the center of this 2x2 block.
   - This strategy ensures that the second player's moves are never inside the convex hull of the marked points, maintaining the convex hull's properties.
   - Therefore, the second player can always guarantee a win when \( n \) is even.

4. **Counting the Values of \( n \) for Which the First Player Wins**:
   - We need to count the number of odd integers between 1 and 2013 (inclusive).
   - The sequence of odd numbers between 1 and 2013 is \( 1, 3, 5, \ldots, 2013 \).
   - This is an arithmetic sequence with the first term \( a_1 = 1 \) and common difference \( d = 2 \).
   - The \( k \)-th term of this sequence is given by \( a_k = 1 + (k-1) \cdot 2 = 2k - 1 \).
   - Setting \( 2k - 1 = 2013 \), we solve for \( k \):
     \[
     2k - 1 = 2013 \implies 2k = 2014 \implies k = 1007
     \]
   - Therefore, there are 1007 odd integers between 1 and 2013.

Conclusion:
The number of values of \( n \) between 1 and 2013 for which the first player can guarantee a win is \( \boxed{1007} \).