Recall that a number is divisible by $11$ exactly if its alternating digit sum $S=d_1-d_2+d_3-d_4...$ is.
From this we immediately glean that a number with the required property must have the same digit in all even and the same digit in all odd positions. Indeed, choose any position and compute the alternating digit sum leaving out that value (without altering the signs of the other terms) $S_{\widehat k}=S+(-1)^kd_k$. Because we are allowed to replace the left out place with any digit, none of $S_{\widehat k},S_{\widehat k}-(-1)^k,S_{\widehat k}-(-1)^k\times 2,\ldots,S_{\widehat k}-(-1)^k\times 9$
can be divisible by $11$. As these are 10 consecutive numbers, the previous and next must be multiples of $11$. I.e. $S_{\widehat k}+(-1)^k \equiv 0 \mod {11}$, from which the assertion follows.
Call those the even and odd digits $d_e,d_o$. Similarly, write $S_{\widehat e},S_{\widehat o}$ for the corresponding omissions. Then $S_{\widehat e} \equiv -1 \mod {11}$ and
$S_{\widehat o} \equiv 1 \mod {11}$. Taking the difference yields $9 \equiv -2 \equiv S_{\widehat e} - S_{\widehat o} \equiv d_e + d_o \mod {11}$.
Let $2n$ be the number of digits. Then $0 \equiv S_{\widehat o} - 1 \equiv (n-1)d_o - nd_e - 1 \equiv (n-1)d_o + nd_o + 2n - 1 \equiv (2n-1)(d_o + 1) \mod {11}$. It follows that $2n-1$ is a multiple of $11$ because $d_o+1$ cannot ($1 \le d_o+1 \le 10$).
The smallest number with the property is therefore
$181818181818$.
