This problem actually arose as a question in the real world (see the paragraph "Origin of the problem" below). 


Let $\mathbb{N}$ denote the set of the positive integers and let $n\in\mathbb{N}$. If $w:\{1,\ldots,n\}\to \mathbb{N}$ is a function (the letter $w$ stands for "weight function") and $S\subseteq \{1,\ldots,n\}$ we say that the *score* of $S$ with respect to $w$ is $$\text{sc}\_w(S) = \sum\_{k\in S}w(k).$$ Suppose that ${\cal A}$ is a set of subsets of $\{1,\ldots,n\}$ such that no member of ${\cal A}$ is contained in another member of ${\cal A}$, and let $a=|{\cal A}|$. 


**Question.** Given a bijective map $\varphi:\{1,\ldots,a\}\to{\cal A}$, is it possible to find a weight function $w:\{1,\ldots,n\}\to \mathbb{N}$ such that for all $k\in\{1,\ldots,a-1\}$ we have $$\text{sc}\_w(\varphi(k)) > \text{sc}\_w(\varphi(k+1)) ?$$


(In more intuitive but less exact terms, is it possible to achieve any ranking of the members of ${\cal A}$ by cleverly choosing the weight function $w:\{1,\ldots,n\}\to\mathbb{N}$?)


**Origin of the problem.** A friend of mine is a math teacher and he noted that in the most recent exam, his students all had solved a different set of problems, such that no student had solved a subset of problems that another student had solved. Every problem gives a certain amount of points. For simplicity he assumed that the students get all points if they solved a certain problem, and no points otherwise. The question he asked me was whether he could achieve any ranking of his students by re-assigning the points to the problems.



Achieving every possible ranking by rearranging weights
The following should give an example where you can't achieve every ranking.


Let $n=4$ and let $\mathcal{A} = \{ (1,2),(1,3),(1,4),(2,3),(2,4),(3,4)\} = [4]^{(2)}$.


I claim that we can't achieve any ranking which starts $(1,2) > (3,4) > \ldots$.


Indeed, since $sc\_w(1,2) > sc\_w(3,4)$ if follows that $\max \{ w(1),w(2) \} > \min \{ w(3),w(4)\}$ and so by symmetry we may assume that $w(1) > w(4)$.


But then $sc\_w(1,3) > sc\_w(3,4)$, contradicting our assumption that only $(1,2)$ was better than $(3,4)$.


