jmc

Recall that $a\equiv 3(\mathrm{mod}7)$ if and only if $a-3$ is divisible by 7. Subtracting 3 from every element in the list gives $$8249,476-7012,000,000-6$$By dividing, we can see that 82 and $-6$ are not divisible by 7, whereas $-70$ and $49,476$ are divisible by 7. To see that $12,000,000$ is not divisible by 7, note that its prime factorization is $(12)(10^6)=(2^2\cdot 3)(2\cdot 5{)}^6=2^8\cdot 3\cdot 5^6$. So, after striking off the numbers which are congruent to 3 (mod 7), the original list becomes $$85\cancel{49,479}\cancel{-67}12,000,003-3$$The sum of the remaining integers is $\boxed{12,000,085}$.