Suppose that the original six integers are $x, x+1, x+2, x+3, x+4$, and $x+5$. Suppose also that the integer that was erased is $x+a$, where $a$ is $0,1,2,3,4$, or 5. The sum of the integers left is $(x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5))-(x+a)$. Therefore, $5(x+3)=2012+a$. Since the left side is an integer that is divisible by 5, then the right side is an integer that is divisible by 5. Since $a$ is $0,1,2,3,4$, or 5 and $2012+a$ is divisible by 5, then $a$ must equal 3. Thus, $5(x+3)=2015$ or $x+3=403$ and so $x=400$. Finally, the integer that was erased is $x+a=400+3=403$. The sum of its digits is $4+0+3=7$.