(*For my easy understanding, let me rewrite the question. The author should feel free to remove my edit or... accept it; I am leaving the original formulation at the end intact*).

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**REFORMULATION**

An ordered pair $(A\ B)$ of subsets of the ring of integers $\mathbf Z$ is called globally ordered if $\ \Leftarrow:\Rightarrow\ \ a \le b\ $ for every $a\in A$ and $b\in B$.

Let $k>1$ be an arbitrary natural number. Let $\ S\ $ be a finite family of k-element subsets of $\mathbf Z,\,$ each two of which have a non-empty intersection. **Is it true that** if no two members of $S$ form a globally ordered pair then it's possible to color $\mathbf Z$ in such a way that every member of $S$ is bicolored **?**

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*PS. I'd like to ask the author to be a bit more detailed and explicit about the definitions related to the coloring.*

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**THE ORIGINAL QUESTION:**

Assume we have a set of numbers $\{ 1, 2, ..., n \}$ and a set of subsets $\{ M_1, ..., M_s \}$, such that $|M_i| = k$ and $ \forall i,j\ (i\ne j\implies|M_i \cap M_j| = 1)$.

Let's define **2-chain** -- it is two subsets, $M_i, M_j$ with one common element - $x$ and for every element $y \in M_i$ true that $y < x$, and and for every element $z \in M_j$ true that $z > x$. It's increasing **2-chain** and in the same way we could define decreasing **2-chain**.

We make a random permutation of numbers.

Proof that if there is exist such permutation that there is no **2-chains**, that it's possible to colour all numbers in a way that all $M_i$ will be bicolored.

I think that we should use Lovász local lemma, but I can't build proper probability space.