"The sum of even numbers between 2m and 2n" is 2m+(2m+2)+.........+(2n-2)+2n=(2n+2m)(n-m+1)/2=(n+m)(n-m+1) "The sum of odd numbers between 2m and 2n" is (2m+1)+(2m+3)+....+(2n-1)=((2n-1)+(2m+1))(n-m)/2=(n+m)(n-m) (n+m)(n-m+1)=(n+m)(n-m)+100 (n+m)(n-m+1)-(n+m)(n-m)=100 (n+m)((n-m+1)-(n-m))=100 n+m=100->(m,n)=(1,99)........(49,51) so the answer is 49

The sum of the even numbers between 2m and 2n is given by: 2m + (2m+2) + (2m+4) +... + 2n = (m+n)(n-m+1). The sum of the odd numbers between 2m and 2n is given by: (2m+1) + (2m+3) + (2m+5) +... + (2n-1) = (m+n)(n-m). From the problem's condition, we know: (m+n)(n-m+1) = (m+n)(n-m) + 100. By simplifying this equation, we find that: m + n = 100. Since m < n , the number of valid pairs (m, n) that satisfy this condition is 49. Thus, the total number of solutions is 49.