Let there be $n$ elements in the set. Let $f_{i,n}$ be the number on position $i$ after $n$ steps. If there are always 2 elements such that one divides another, then there are 2 positions $i,j$ such that there are infinitely many $n$ such that $f_{i,n} \mid f_{j,n}.$ But sequence $\frac{ f_{j,n} }{ f_{i,n} }$ is strictly decreasing so we got strictly decreasing sequence of positive integers, contradiction.

if $ k , t \in {M_{n-1}} $ and $ b_n k + 1 | b_n t + 1 $ so we have $ b_n k + 1 | k-t $ so we should have $ max ({r\in M_{n-1} }) - min ({ r \in M_{n-1} }) \geq min ({ r\in M_{n} }) $ $ (1) $ we can prove easily that $ b_n \geq n-2 $ and $ max ({r\in M_n }) - min ({r\in M_n }) = b_n ...b_2 (M-m) $ ( $ M , m $ are $max$ and $min$ in $ M_1 $ ) and $ min ({ r\in M_n }) \geq b_n ... b_2 m + b_{n-1}...b_2 $ from $ (1) $ we should have $ b_2 ... b_{n-1} (M - m -1) \geq b_2 ... b_n m \leftrightarrow {\frac {M - m - 1}{m} \geq b_n} $ and because $ b_n \geq n-2 $ the problem is finished $ \blacksquare $