Let $n$ be a positive integer and let $G$ be the set of points $(x, y)$ in the plane such that $x$ and $y$ are integers with $1 \leq x, y \leq n$. A subset of $G$ is called [i]parallelogram-free[/i] if it does not contains four non-collinear points, which are the vertices of a parallelogram. What is the largest number of elements a parallelogram-free subset of $G$ can have?
1. **Define the set \( G \) and the problem constraints:**
   Let \( G \) be the set of points \((x, y)\) in the plane such that \( x \) and \( y \) are integers with \( 1 \leq x, y \leq n \). We need to find the largest number of elements in a subset of \( G \) that does not contain four non-collinear points forming a parallelogram.

2. **Construct a subset with \( 2n-1 \) points:**
   Consider the points \((1, i)\) and \((i, 1)\) for all \( i \in \{1, 2, 3, \ldots, n\} \). This gives us \( n \) points on the first row and \( n \) points on the first column, but the point \((1, 1)\) is counted twice. Therefore, we have \( 2n - 1 \) points in total.

3. **Verify that no four points form a parallelogram:**
   To form a parallelogram, we need four points \((x_1, y_1)\), \((x_2, y_2)\), \((x_3, y_3)\), and \((x_4, y_4)\) such that the midpoints of the diagonals coincide. However, in our construction, any set of four points will either be collinear or will not satisfy the parallelogram condition due to the specific arrangement of points.

4. **Assume a larger subset and derive a contradiction:**
   Assume there exists a parallelogram-free subset of \( G \) with \( 2n \) points. We will show that this leads to a contradiction using the Pigeonhole Principle.

5. **Consider horizontal distances:**
   Each horizontal line can have at most \( n-1 \) different distances between points. If we place 2 points on each of the \( n \) horizontal lines, we have \( n \) distances. By the Pigeonhole Principle, at least two of these distances must be the same.

6. **Apply the Pigeonhole Principle:**
   Moving a point from one horizontal line to another does not reduce the number of different horizontal distances. Thus, we will still have at least two horizontal distances of the same length in different lines, forming a parallelogram with these four points. This contradicts the assumption that the subset is parallelogram-free.

7. **Conclusion:**
   We have shown that a parallelogram-free subset of \( G \) cannot have more than \( 2n-1 \) points. Therefore, the largest number of elements in a parallelogram-free subset of \( G \) is \( 2n-1 \).

\(\blacksquare\)

The final answer is \( \boxed{ 2n-1 } \).