Let x represent the smallest integer in any case of consecutive
integers with sum 2015.

The sum of n consecutive positive integers beginning with x, 
using the sum formula for an arithmetic sequence 



with a1=x and d=1



Setting that equal to 2015



Multiplying both sides by 2 to clear the fraction
and simplifying:









, the smallest integer.

From ,

we see that n, the number of terms, and 2x+n-1 make up a factor 
pair of 4030.  

We can factor 4030 into a pair of factors the following 8 ways:


n×(2x+n-1) = 4030, x = (4030+n-n^2)/(2n) = smallest integer
----------------------------------------
1×4030             x = 2015  <--ignore since only 1 integer, we need 2 or more  
2×2015             x = 1007                 
5×806              x =  401
10×403             x =  197
13×310             x =  149
26×155             x =   65
31×130             x =   50  
62×65              x =    2
---------------------------
               total = 1871  



The total of the smallest integers x is 1871

Answer: 1841

Edwin