I don't think you can get a solution to your problem!!!. Why? Because no matter what values of x you have between 19001 and 20000, the numerator, in your summation, will always be smaller by a significant amount than the denominator, so your answer will be ~10^-5, rather than .95 that you want.
Your answer will be out by a factor of about 100,000. Sorry, that is the best I can do.
When I summed ALL the values of x from x=19,002 to 19,999 using Wolfram/Alpha computing engine, I got this result which is EXACTLY the same as denominator of 40,000nCr20,000, which renders the result =1, which appears to meet your criteria. Here is the result:
sum_(x=19002)^19999 binomial(39001, x) binomial(999, 20000-x)~~6.320244956184859009513094953017778552721×10^12038 [This is the numerator]..............(1)
6.320244956184859009513094953017778552721×10^12038 [This is the denominator]...........(2)
Divide 1/2=1 EXACTLY.
