Question 1163462
There are 100 potential integers, 50 even and 50 odd.
Two of the odd are not counted (1 and 3)
One of the even is not counted (2).
The way to line them up is:
4+6+8+10+...+100. there are 49 even numbers in the series
---5+7+9+...+99. There are 48 odd numbers in the series
The even is greater with a larger number for the last 48 in the series and a leading number of 4 as well.

There are 48 pairs where E is 1 greater and a 4 that is not paired, so 52 greater.
if you want the sum (not the question)
(n/2)*(2a+(n-1)d)
(49/2)*(8+48*2)=2548 for the even

(48/2)*(10+47*2)=2496 for the odd
They sum to 5044
add the missing numbers 1,2,3 and the sum is 5050, the sum of the first hundred integers
n(n+1)/2=100*101/2=5050