We’ll prove by induction that if there are $2n$ ones, and $2n$ twos, we can perform some operations and get a multiple of $11$, where $n$ is a positive integer. Define $poles$ to be a pair of digits that are used to perform the problem’s operation. Base case $n=1$ is easy, as $1122$, $1221$, $2112$ and $2211$ are all multiples of $11$, and $1212$ and $2121$ can be rearranged into $1122$ and $2211$ by taking the endpoints as poles. Assume it works for $n=k$. Let $N$ be a number with $2k+2$ ones and $2k+2$ twos, $B$ the number formed by the last $4$ digits of $N$, and $A$ the remaining number. Define the $middle$ of $A$ as the number formed by removing the first digit of $A$. Clearly, if at some point $B$ has $2$ ones and $2$ twos, we are done by the induction hypothesis. Assume $WLOG$ that there are more twos than ones in $B$. Therefore, $A$ has at least $3$ ones and we have three cases: Case 1: $B = 2122$, $2212$ or $2221$. There are clearly $2k+1$ ones in $A$, and they can’t all be at the leftmost endpoint. Therefore, take the digit to the left of the rightmost one in $A$ and the second digit of $B$ as poles, perform the operation, and $B$ will have $2$ ones and $2$ twos, done. Case 2: $B = 1222$ There are clearly $2k+1$ ones in $A$, and they can’t all be at the leftmost endpoint; and there are $2k-1$ twos in $A$. If there are no pairs of consecutive ones, all ones in $A$ must have a two or nothing to their left, but this is impossible, as this means there are at least $2k$ twos in $A$, but we know there are only $2k-1$ of them. Therefore, there must be a pair of consecutive ones somewhere in $A$. If there are $2$ ones in the middle of $A$ that are consecutive, then we are done, as we take the digit to the left of that pair as a pole and the second $2$ in $B$ as the other one, perform the operation, and $B$ will have $2$ ones and $2$ twos, done. If there is no such pair in the middle of $A$, this implies that $A$ must start with $2$ ones. As there are no other pairs of ones in $A$, the other ones in $A$ must have a two to their left. As these ones appear $2k-1$ times in $A$, and so do all twos in $A$, $A = 11212121 \dots 21$. But then $N = 11212121 \dots 211222$, and we get a pair of ones that is not at the leftmost point, which consists of $A$’s rightmost one and $B$’s leftmost one (and clearly, $A$’s rightmost one is not in the first position). We perform the operation by taking $A$ rightmost two and the third digit of $B$ as poles, and $B$ will have $2$ ones and $2$ twos, done. Case 3: $B = 2222$ There are clearly $2k+2$ ones in $A$ and $2k-2$ twos in $A$. If there are no pairs of consecutive ones, all ones in $A$ must have a two or nothing to their left, but this is impossible, as this means there are at least $2k+1$ twos in $A$, but we know there are only $2k-2$ of them. Then there is a pair of consecutive ones somewhere in $A$. If such pair is in the middle of $A$, we take the digit to the left of such pair as a pole, and the third two in $B$ as poles, and we are done, as $B$ will have $2$ ones and $2$ twos. If no such pair exists in $A$, then $A$ must start with $2$ ones, and the $2k$ remaining ones must have a two to their left, which is a contradiction, as there are only $2k-2$ twos. Therefore, take $n = 5$ and the problem is solved.