If $ 11\mid\overline{abcd}$, then we must have $a+c=b+d$, $a+c=b+d+11$, or $a+c=b+d-11$. (Remember $0\le a,b,c,d\le 9$ and $0<a$) Case 1:$a+c=b+d$ Then we have $a+b+c+d=2(b+d)=d^2$ or $2b+1=(d-1)^2$. From here, one easily finds the pairs that satisfy the equation is $(b,d)=(4,4),(0,2)$. For $(b,d)=(4,4)$, we have $a+c=8$ and $(a,c)=(1,7),(2,6),(3,5),(4,4),(5,3),(6,2),(7,1),(8,0)$. The only case where $7\mid \overline{ac}$ is when $\overline{ac}=35$. Thus we have $3454$ as one answer. For $(b,d)=(0,2)$, we have $(a,c)=(1,1),(2,0)$ of which none have $7\mid \overline{ac}$. Thus the only number in this case is $3454$. Case 2:$a+c=b+d+11$ Now we get $a+b+c+d=2(b+d)+11=d^2$ or $2(b+6)=(d-1)^2$. We then find the only pair $(b,d)=(2,5)$. Then $a+c=11+7=18$ and $(a,c)=(9,9)$. But $7\nmid 99$ so there are no such numbers in this case. Case 3:$a+c=b+d-11$ Then $a+b+c+d=2(b+d)-11=d^2$ or $2(b-5)=d^2$. Then after testing values, we get $(b,d)=(7,3),(5,1)$. But for both cases we have $b+d<11$ and so $a+c<0$, a contradiction. Thus there are no such numbers in this case. Thus the only number is $\boxed{3454}$.