jmc

Since $11065,11067,11069,\dots ,11077$ are $7$ consecutive odd integers, they include exactly one integer from each of the residue classes $1,3,5,7,9,11,13(\mathrm{mod}14)$ (not necessarily in that order). Therefore, their sum is congruent $(\mathrm{mod}14)$ to $1+3+5+7+9+11+13=49$. The remainder of this sum $(\mathrm{mod}14)$ is $\boxed{7}$.