This problem is not actually that difficult. Firstly, we note that if $n$ is even, then $A_n$ is simply $3^n$ ones in a row, and that if $n$ is odd, then $A_n$ is simply $3^n$ repetitions of the string $100$. This implies that $a_n = 0$ iff $n$ is odd and $n-1$ is not a multiple of $3$. Then we can pick $m=6$. For every positive integer $l$, there is exactly one integer $k\leq m$ such that $l+k\equiv 0\pmod{6}$, and for such a $k$ the stated property is satisfied.

duck_master wrote: Firstly, we note that if $n$ is even, then $A_n$ is simply $3^n$ ones in a row, and that if $n$ is odd, then $A_n$ is simply $3^n$ repetitions of the string $100$. I think the description about the operation caused misunderstanding. $A_n$ is a sequence starts with $100,10011,10011100100,\dots$, since the two changes--$1$ into $100$ and $0$ into $1$--occur at the same time, not one by one. Added the first few terms of $A_n$ to the statement.