jmc

For an integer $abcde$ to be divisible by $11$ then $a-b+c-d+e$ is divisible by $11.$

We start with the case where $(a+c+e)-(b+d)=0.$ Then, $a+c+e=b+d.$ Since we have a palindrome, we must have $a=e$ and $b=d,$ meaning that $2a+c=2b.$ We must have that $a$ and $e$ must be at least $1,$ so then we can let $b$ and $d$ also be 1 and $c$ be zero. So the smallest such five-digit palindrome is $11011.$

Then, we investigate the case where $(a+c+e)-(b+d)=11.$ Then, $a+c+e=b+d+11,$ and $a=e$ and $b=d,$ so $2a+c=11+2b.$ We see that we can let $a$ be 1 and $c$ be 9, then $b=0,$ and we have the palindrome $10901.$

Finally, $(a+c+e)-(b+d)=-11.$ Then, $2a+c=2b-11.$ We check if $a=1$ has any solutions. We get $2+c=2b-11,$ so $c-2b=-9.$ Then, we can see that there are no solutions for $b=0$, since then we would have $c=-9.$ Since we already found $10901,$ we do not need to check for any $b$ greater than $0$, so we see that our solution is $\boxed{10901}.$