**330**

Solution:

Sum of all numbers from 1 to n is given by n(n+1)/2 i.e. 100*(100+1)/2 = 5050

6 comes in 'units' place 10 times in the numbers 1 to 100 i.e. 6,16...96

6 comes in 'tens' place 10 times in the numbers 1 to 100 i.e. 60,61...69

sum of these 6's is 10(6*10)+10(6*1)=600+60=660

if all 6's are replaced by 9

then sum of those 9's will be 10(9*10)+10(9*1)=990

now algebraic sum will be 5050-660+990= 5380

Difference: 5380 - 5050 = 330