WOW, very nice problem: By Minsky theorem, if there is no chain of lenght $m+1$ in the division poset, then it is the union of $m$ antichains. Now let $C_1,C_2...C_m$ be this antichains, For every elemento of a chain $a$, map it to $[2^{m-1}n+1,2^mn]$ in the following way: choose the integer $t$ such that $2^ta$ belongs to that interval. Now, numbers in an antichain go to different numbers in this map, and there can be overlaps between numbers of different antichains. Suppose the max power of 2 dividing $k$ in $[2^{m-1}n+1,2^mn]$ is $t$, then we can get to $k$ from $t+1$ numbers, or at most $m$, because, there cannot be two numbers in the same antichain with same image. So we have at most: $2^{m-2}n+2^{m-3}2n+2^{m-4}3n...2^{m-k}(k-1)n...2(m-2)n+2(m-1)n+mn=(2^m-1)n$ possible images counting repetitions, so if we have our Poset partitioned in $m$ antichains, then it has at most $(2^{m}-1)n$ elements, contradiction. Minsky, or Mirsky theorem is a kind of dilworth dual theorem for Posets. this is a nice generalization of the erdos problem which is the case $m=1$. Great Problem

i think it can be done using induction after $m$ $m=1$ well known now let $U=\left\{1,2,\ldots,2^{m}n\right\}$ and $V=\left\{2^{m}n+1,\ldots,2^{m+1}n\right\}$. and $S\subseteq(U\cup V)$ our subset. we make a partition of $S$ : $S_{u}\subseteq U$ $S_{v}\subseteq V$. $S_{u}\cap S_{v}=\emptyset$. Now, $|V|=2^{m}n$, so let $|S_{v}|=2^{m}n-k$. (because $|S_{v}|\leq|V|$). Then $|S_{u}|=|S|-|S_{v}|=(2^{m+1}-1)n+1-2^{m}n+k=(2^{m}-1)n+1+k$. (*) I will call "end of p-chain" a number which is the end of a chain of type asked by the problem, of length $p$. For example, in the chain $1|2, 2|4, 4|8$, $8$ is a "end of a 4-chain". Using the induction hypothesis, and (*) above, it follows that $S_{u}$ has $k+1$ "ends of (m+1)-chains". This is because, we find the first "end", let it be $e_{1}$, then we look at the set $S_{u}-e_{1}$, etc.. Let the $k+1$ "ends of (m+1)-chains" be $e_{1},e_{2},\ldots,e_{k+1}$. We take the set $Z=\left\{e_{1},e_{2},\ldots,e_{k+1}\right\}\cup S_{v}$. Also let $t=2^{m}n$. The set $Z$ has exactly $t+1$ elements, and all elements are no bigger than $2t$. Then, it is the well known problem (case $m=1$), so there must be two which divide each other in $Z$. Let this numbers be $a$ and $b$, and $a|b$. If $a\in S_{v}$, then $2a>2t$, and because $a|b$, we get $b\geq 2a>2t$, contradiction. So $a$ must be a "end of (m+1)-chain". we take this chain and complete it with $b$, and there is the $m+2$ chain. this finishes the induction

By Mirsky theorem, we have to prove that the minimal number of anti chains which has union as $S$ is $\ge m+1$ We will show that even $m$ anti chains can not cover $S$. There are total $2^{m-1} n$ odd numbers in the set $\left \{ 1, \ldots, 2^m \cdot n \right \} $.Write each number as $2^k \cdot b$, where $b$ is odd. All anti chain will contain distinct odd numbers. (otherwise one will divide the other). We can fill all the anti chains with all odd numbers. So there are total $2^{m-1} n \cdot m$ numbers .(not distinct). Now we remove the numbers that are not distinct. An odd number from the set $\left \{ 2^{m-k}n + 1, 2^{m-k+1}n - 1 \right \}$ can contribute in at most $ k $ distinct antichains (here $1 \le k \le m-1$ ) and there are $2^{m-k-1}n$ odd numbers in that interval.So there are $(m-k) 2^{m-k-1}n$ entries which are same. (has already appeared). We remove them. So number of distinct entries is $2^{m-1}mn - \sum_{k=1}^{m-1}n (m-k)2^{m-k-1} = (2^m-1)n $. Which is a contradiction since $|S| = (2^m-1)n + 1$. QED

Use induction on M with this Statement: A: $\{1,2,3,\ldots, 2^mn\}$ If we delet K elements from A (k<n) then we have N-K subset of A :(B1,B2.....Bn-k) such that each of them has M+1 elements like a1,a2,....,am+1 such that $a_{k-1} \mid a_{k}$ and if F(Bi)=greatest member of Bi then all of F(B1),F(B2),....,F(Bn-k) are different.