Prair has a box with some combination of red and green balls.  If she randomly draws two balls out of the box (without replacement), the probability of drawing two balls of the same color is equal to the probability of drawing two balls of different colors!  How many possible values between $200$ and $1000$ are there for the total number of balls in the box?

[i]Proposed by Andrew Wu[/i]
1. Let the number of red balls be \( x \) and the number of green balls be \( y \). The total number of balls is \( x + y \).

2. The probability of drawing two balls of the same color is:
   \[
   \frac{\binom{x}{2} + \binom{y}{2}}{\binom{x+y}{2}} = \frac{\frac{x(x-1)}{2} + \frac{y(y-1)}{2}}{\frac{(x+y)(x+y-1)}{2}} = \frac{x(x-1) + y(y-1)}{(x+y)(x+y-1)}
   \]

3. The probability of drawing two balls of different colors is:
   \[
   \frac{xy + yx}{\binom{x+y}{2}} = \frac{2xy}{(x+y)(x+y-1)}
   \]

4. Given that these probabilities are equal, we set up the equation:
   \[
   \frac{x(x-1) + y(y-1)}{(x+y)(x+y-1)} = \frac{2xy}{(x+y)(x+y-1)}
   \]

5. Simplifying, we get:
   \[
   x(x-1) + y(y-1) = 2xy
   \]

6. Expanding and rearranging terms:
   \[
   x^2 - x + y^2 - y = 2xy
   \]
   \[
   x^2 + y^2 - x - y = 2xy
   \]
   \[
   x^2 + y^2 - 2xy = x + y
   \]
   \[
   (x - y)^2 = x + y
   \]

7. Let \( a = x + y \). Then:
   \[
   (x - y)^2 = a
   \]

8. Since \( (x - y)^2 = a \), \( a \) must be a perfect square. Let \( a = k^2 \) for some integer \( k \).

9. We need to find the number of perfect squares between 200 and 1000. We solve for \( k \):
   \[
   200 \leq k^2 \leq 1000
   \]
   \[
   \sqrt{200} \leq k \leq \sqrt{1000}
   \]
   \[
   14.14 \leq k \leq 31.62
   \]

10. Since \( k \) must be an integer, the possible values of \( k \) are from 15 to 31 inclusive.

11. The number of integers from 15 to 31 is:
    \[
    31 - 15 + 1 = 17
    \]

The final answer is \(\boxed{17}\)