jmc

Since $7438,7439,\dots ,7445$ are $8$ consecutive integers, they include exactly one integer from each residue class $(\mathrm{mod}8)$. Therefore, their sum is congruent $(\mathrm{mod}8)$ to $0+1+2+3+4+5+6+7=28$. The remainder of this sum $(\mathrm{mod}8)$ is $\boxed{4}$.