Fifty-Fifty

Suppose there are $a$ red marbles and $b$ white marbles. The number of ways to draw two different-colored marbles is $ab + ba = 2ab$ . The number of ways to draw two same-colored marbles is $a(a-1) + b(b-1)$ . Setting these equal and moving some terms around, we get $(a-b)^2 = a+b$ . So $a+b$ , the number of marbles, has to be a square. The largest square less than $1000$ is $\fbox{961}$ . It's not hard to check that $a =496, b=465$ satisfies the conditions of the problem.

Let $r$ be the number of red marbles.Then the number of white marbles is $n-r$ .The number of ways we can draw two marbles of the same color is $r(r-1)+(n-r)(n-r-1)$ while the number of ways we can draw two marbles of different color is $r(n-r)+(n-r)r$ .These numbers have to be equal for the probability to be the same so we have $r(r-1)+n(n-1)=2r(n-r) \Leftrightarrow 4r^2-4nr+n^2-n=0$ .Solving for r we get $r=\frac{n-\sqrt{n}}{2}$ or $r=\frac{n+\sqrt{n}}{2}$ .Since $r$ is an integer $n$ must be a perfect square.The largest perfect square less than 1000 is 961.Substituting in our last equations to check if the solutions are integral we get $r=465$ and $r=496$ so the number of white marbles is respectively 496 and 465.

For any $n$ square, there is a solution with $a=\frac{n-\sqrt{n}}{2}$ and $b=\frac{n+\sqrt{n}}{2}$ . It's easy to check they're always an integer, and that they are between $0$ and $n$ .