Given a grid rectangle of size $2010 \times  1340$. A grid point is called [i]fair [/i] if the $2$ axis-parallel lines passing through it from the upper left and lower right corners of the large rectangle cut out a rectangle of equal area (such a point is shown in the figure). How many fair grid points lie inside the rectangle?
[img]https://cdn.artofproblemsolving.com/attachments/1/b/21d4fb47c94b774994ac1c3aae7690bb98c7ae.png[/img]
1. Let \( x \) and \( y \) be the side lengths of the top right rectangle. The area of the top right rectangle is \( xy \). The area of the bottom left rectangle is \((2010 - x)(1340 - y)\).

2. Since the areas of these two rectangles are equal, we have:
   \[
   xy = (2010 - x)(1340 - y)
   \]

3. Expanding and simplifying the equation:
   \[
   xy = 2010 \cdot 1340 - 2010y - 1340x + xy
   \]
   \[
   0 = 2010 \cdot 1340 - 2010y - 1340x
   \]
   \[
   2010y + 1340x = 2010 \cdot 1340
   \]

4. Dividing the entire equation by 670 (the greatest common divisor of 2010 and 1340):
   \[
   \frac{2010y}{670} + \frac{1340x}{670} = \frac{2010 \cdot 1340}{670}
   \]
   \[
   3y + 2x = 4020
   \]

5. We need to find the number of positive integer solutions to the equation \( 3y + 2x = 4020 \).

6. Solving for \( y \) in terms of \( x \):
   \[
   3y = 4020 - 2x
   \]
   \[
   y = \frac{4020 - 2x}{3}
   \]

7. For \( y \) to be an integer, \( 4020 - 2x \) must be divisible by 3. Since \( 4020 \equiv 0 \pmod{3} \), \( 2x \equiv 0 \pmod{3} \). Therefore, \( x \) must be a multiple of 3.

8. Let \( x = 3k \) for some integer \( k \). Substituting \( x = 3k \) into the equation:
   \[
   y = \frac{4020 - 2(3k)}{3}
   \]
   \[
   y = \frac{4020 - 6k}{3}
   \]
   \[
   y = 1340 - 2k
   \]

9. We need \( y \) to be a positive integer, so:
   \[
   1340 - 2k > 0
   \]
   \[
   1340 > 2k
   \]
   \[
   k < 670
   \]

10. Additionally, \( x = 3k \) must be a positive integer, so:
    \[
    k > 0
    \]

11. Therefore, \( k \) ranges from 1 to 669, inclusive. This gives us 669 possible values for \( k \).

The final answer is \(\boxed{669}\).