TITLE: A problem with 26 distinct positive integers
QUESTION [13 upvotes]: I am trying to solve the following problem. 
Assume that we are given 26 distinct positive integers.
Show that either there exist 6 of them $x_1<x_2<x_3<x_4<x_5<x_6$, with $x_1$ dividing $x_2$,
$x_2$ dividing $x_3$, $x_3$ dividing $x_4$, $x_4$ dividing $x_5$ and $x_5$ dividing $x_6$ or there exist six of them, such that none of them divides another one of these six.
A possibly good start is to assume that, in every six of these numbers, there exists at least one dividing another one of the same six.
Update. I have found a solution of the problem (for 17 numbers though) in a Russian site. As unbelievable as it may sound, this problem was a question in a 1983 Soviet Mathematics contest (Турниры Городов) for student of 7-8 grades!
I am presenting the solution I found in that site below as an answer, 
and it is generalised for $n^2+1$ distinct integers, where we show that either there exist
$n+1$ of them dividing each other or none diving none else.

REPLY [2 votes]: I have found this solution (for 17 numbers though) in a Russian site, as this problem was a question in a 1983 Soviet Mathematics contest (Турниры Городов) for student of 7-8 classes!
Solution. We shall attach on each of our numbers
$$
x_1<x_2<\cdots<x_{26},
$$
an index next to their subscript in the following fashion:
$x_1$ becomes $x_{1,1}$.
If $x_1$ divides $x_2$, then $x_2$ becomes $x_{2,2}$, otherwise it becomes $x_{2,1}$.
In general, if we have attached indices to $x_1,\ldots,x_k$, then the index of $x_{k+1}$ will be $1$ is none of the $x_1,\ldots,x_k$ divides $x_{k+1}$, or it will become $i+1$, if $i$ is the largest index of all the numbers which divide $x_{k+1}$.
If the index of some of the $x_j$'s is at least $6$ then we have have a sequence
$$
x_{k_1,1} \mid x_{k_2,2} \mid x_{k_3,3} \mid x_{k_4,4}
\mid x_{k_5,5}\mid x_{k_6,6}.
$$ 
In all the indices are less or equal to $5$, then some number in $\{1,2,3,4,5\}$, say $j$, is necessarily the index of at least $6$ numbers, i.e.
$$
x_{k_1,j} < x_{k_2,j} < x_{k_3,j} < x_{k_4,j} < x_{k_5,j} < x_{k_6,j},
$$
which means that none of the above $6$ numbers divide another one of them.
Note. Τhis can be generalised to finding a chain of $k$ numbers, fully ordered by division, or a subset of $\ell$ numbers, with none of them dividing another one, among $m$ distinct positive integers, whenever $(k-1)(\ell-1)\le m-1$.