$2018=2\cdot 1009$. Let $2k+1$ denote the number of digits of $A$ $( k\in \mathbb{N})$. We can take a triple of positive integers $( X,Y,c) \in \mathbb{Z}^{3}$ such that $10^{k-1} \leq X,Y< 10^{k}$, $0\leq c\leq 9$, and $A=X\cdotp 10^{k+1} +c\cdotp 10^{k} +Y$. Then we have $B=10^{k} X+Y$, of course. Note that both $A$ and $B$ are divisible by $2018$ if and only if both $A-B$ and $B$ are divisible by $1009$ and $2\mid A$. So $A$ satisfies the condition if and only if there exists $( X,Y,c,k) \in \mathbb{N}^{4}_{\geq 0}$ such that $10^{k-1} \leq X,Y< 10^{k}$, $0\leq c\leq 9$, $2\mid Y$, and $1009\mid c+9X,Y+10^{k} X$ where $A=X\cdotp 10^{k+1} +c\cdotp 10^{k} +Y$, Suppose that $A$ satisfies the condition. Since $1009\mid c+9X$, we must have $k\geq 3$. Assume that $k=3$. Since $0\equiv Y+10^{3} X\equiv Y-9X\equiv Y+c\pmod{1009}$, we have $1009\leq Y+c\leq 999+9=1008$. This is clearly impossible. So we must have $k\geq 4$. Let $d=( c+9X) /1009\in \mathbb{N}$. Since $c+9X\geq 9000$, we must have $d\geq 9$. Note that $d=9$ if and only if $X=1009$ and $c=0$. If $d=9$, then $1009\mid Y$ because $X=1009$ and $1009\mid Y+10^{k} X$. Therefore, we will get the minimum value of $A$ when $X=1009,Y=2018$. Hence the answer is $\boxed{A=100902018}$.