The minimun of $|A\triangle B|$ is $0,m=2$ and $2m-2,m\ge3$,I think. proof: When $m=k-1\ge3$,suppose that $|B|\ge 3m-2=3k-5$ ,which is obvious when $m=3$.When $m=k$ 1.if $a_k+a_{k-3}\neq 2a_{k-1}$ $a_k+2a_{k-1} ,2a_k+a_{k-1} $,and at least one of$a_k+2a_{k-3},2a_k+a_{k-3}$can't be denoted as $a_i+2a_j,1\le i,j\le k-1$and they have different values. 2.if $a_k+a_{k-3}=2a_{k-1}$ so$a_k+2a_{k-2}> a_k+a_{k-3}+a_{k-2}=2a_{k-1}+a_{k-2}$ $a_k+2a_{k-1} ,2a_k+a_{k-1},a_k+2a_{k-2} $ can't be denoted as $a_i+2a_j,1\le i,j\le k-1$and they have different values. we get $|B|\ge 3k-5+3=3m-2$. So $|A\triangle B|=|A|+|B|-2|A\cap B|\ge m+3m-2-2m=2m-2$ example:take$a_i=i-2,m\ge3$ and $a_1=-1,a_2=1,m=2$

zsgvivo wrote: The minimun of $|A\triangle B|$ is $0,m=2$ and $2m-2,m\ge3$,I think. proof: When $m=k-1\ge3$,suppose that $|B|\ge 3m-2=3k-5$ ,which is obvious when $m=3$.When $m=k$ 1.if $a_k+a_{k-3}\neq 2a_{k-1}$ $a_k+2a_{k-1} ,2a_k+a_{k-1} $,and at least one of$a_k+2a_{k-3},2a_k+a_{k-3}$can't be denoted as $a_i+2a_j,1\le i,j\le k-1$and they have different values. 2.if $a_k+a_{k-3}=2a_{k-1}$ so$a_k+2a_{k-2}> a_k+a_{k-3}+a_{k-2}=2a_{k-1}+a_{k-2}$ $a_k+2a_{k-1} ,2a_k+a_{k-1},a_k+2a_{k-2} $ can't be denoted as $a_i+2a_j,1\le i,j\le k-1$and they have different values. we get $|B|\ge 3k-5+3=3m-2$. So $|A\triangle B|=|A|+|B|-2|A\cap B|\ge m+3m-2-2m=2m-2$ example:take$a_i=i-2,m\ge3$ and $a_1=-1,a_2=1,m=2$ Isn't the sequence made up of positive integers