Nice question!
First calculate the number of odd integers between 1 and 399 inclusive that comes up to 200 because the number of the even and odd integers are equal in quantity.
So the total number of integers to chose from is 200.
Now r<s this condition restricts the result:
What we can do is to choose r and then select s:
so r can be selected in 1 way for example 1 then s can selected in 199 ways from 3 to 399.
Similarly we can select r as 3 the number of ways of doing so is also 1 but the number of ways that s can be selected is now 198.
So the series becomes:
(1*199) + (1*198) + (1*197) + .... + (1*1)
We can calculate the sum of this series as [(1+199)/2] * 199 = 19900.
So the answer is 19900.

Sir, I notice that there are 200 odd numbers and in between 1 and 10, we have the ten possibilities that meet the criteria. On doing this way, I got only 400 pairs, what do I missing sir?