How many ordered pairs of positive integers (r,s), subject to 1000?r,s?2000, are there such that if r red Chri?

There are 34 ordered pairs.
First observe that any (r,s) solution is also an (s,r) solution as there's nothing specific about either r or s, the problem is fully symmetrical. Therefore we'll establish that there are 17 ordered pairs that satisfy the problem's conditions in which r > s, and will simply swap r and s to obtain the other 17 symmetric solutions (in which s > r).

Basic data:
r > s
1000 <= r <= 2000
1000 <= s <= 2000

Therefore 2000 <= r+s <= 4000

Now on to calculating the probability. Suppose we have our r + s balls ready for placement.
The indicated probability isn't affected at all by our specific procedure of placing the balls, as long as we consider them hidden in a box and drawn at random. So we put them as follows:
1st, we fill the leftmost end.
2nd, we fill the rightmost end
3rd, we fill the new leftmost end
etc etc.
So, the indicated probability is only dependent on the first two placements and can be calculated very easily as follows:

P(both ends have the same colour) =

P(first draw is red AND second draw is red) +
P(first draw is silver AND second draw is silver) =

r/(r+s) * (r-1)/(r+s-1)  +  s/(r+s) * (s-1)/(r+s-1)  =

(r^2 -r + s^2 - s) / (r^2 + s^2 + 2rs -r -s) and this must equal 1/2

It follows that: 2r^2 - 2r + 2s^2 - 2s = r^2 + s^2 + 2rs -r -s or

r^2 + s^2 - 2rs = r+s or

(r-s)^2 = r+s (1)

Let r-s = d so r = s+d (2) (recall that r > s so d >0)

By (1) d^2 = r+s therefore 2000 <= d^2 <= 4000

So 46 <= d <= 63 , since d is an integer.

And 1000 <= s <= 1937, so that r = s + d will not exceed 1937 + 63 = 2000

Now, r = s + d therefore r + s = 2s + d, and thus (1) (r-s)^2 = 2s+d or

d^2 = 2s + d or

s = (d^2 - d) / 2

So, furthermore, (d^2 - d) / 2 must lie between 1000 and 1937
Solving the two quadratic inequalities, we obtain 46 <= d <= 62, which are the acceptable values of d. They are 17 in total.
For each one of them, we get an acceptable value for s = (d^2 - d) / 2, and each pair (s,d) gives an acceptable value of r = s + d. So it's eventually the values of d that determine uniquely the number or (r,s) pairs.

Hence there are 17 acceptable (r,s) pairs in which r > s, therefore there are another 17 acceptable pairs in which s > r, each being a swap of the first 17.

The total number of acceptable pairs is 34.

BTW, the 17 (r,s) , r > s pairs are:

10811035
11281081
11761128
12251176
12751225
13261275
13781326
14311378
14851431
15401485
15961540
16531596
17111653
17701711
18301770
18911830
19531891