Ordering and place in sets 
Given are $2n$ (not necessarily distinct) subsets of the set $\{1,2,\dots,k\}$, with the first $n$ sets containing $1$. For an ordering $\sigma$ of $1,2,\dots,k$, its score is calculated as follows, starting from $0$: Going through $\sigma$, for element $x$ in the $i$th place of $\sigma$, for each set, if $x$ belongs to the set and is the $j$th element of the set to be considered, we add $\frac{1}{ij}$ to the score. Is it true that all $\sigma$ maximizing the score must have at least one element from the first $n$ sets in the first two places?

Example: $n=2,k=4$ and four subsets $\{1,2\},\{1,3\},\{2,3\}$ and $\{2,4\}$. The score of $\sigma=(2,1,3,4)$ is calculated by going through $\sigma$ in order. The element $2$ adds score $\frac{3}{1\times 1}$, the element $1$ adds score $\frac{1}{2\times 1}+\frac{1}{2\times 2}$, the element $3$ adds score $\frac{2}{3\times 2}$ and the element $4$ adds score $\frac{1}{4\times 2}$. This gives total score $\frac{101}{24}$.
For motivation, the score function is election scheme variant, and the question asks whether this function can represent voters who like a common candidate.
 A: A quick computer program throws up a counter-example for $n=2$, $k=4$: take subsets $\{1\}, \{1, 2, 3\}, \{2\}, \{3\}$. The maximum score of $\frac{115}{36}$ is obtained by $(1, 2, 3, 4)$, $(1, 3, 2, 4)$, $(2, 1, 3, 4)$, $(3, 1, 2, 4)$, $(2, 3, 1, 4)$, $(3, 2, 1, 4)$ and the last two fail the condition.
A: Although I could not arrive to answering the question itself, I can at least say: all $\sigma$ maximizing the score must have at least one element from the first $\frac{4}{3}n$ sets in the first two places.
Take $\sigma$ not satisfying the above. In particular, all the elements of the first $n$ sets (including $1$) are at least in 3rd place, and the element $a$ in 1st place is appears in at most $\frac{2}{3}n$ sets: we want to prove then that swapping $1$ and $a$ makes the score grow.
The swapping has no effect on the score coming from the sets in which both elements are contained: in each such set, the parts of the score coming from $1$ and $a$ are exchanged and the score of all the other elements is left untouched (both their position in $\sigma$ and in $S$ are not altered). Obviously, the sets that do not contain either element are also unchanged.
Suppose that $S$ is one of the first $n$ sets: by hypothesis it contains $1$ but not $a$. The score coming from the elements after $1$ (in the original ordering of $\sigma$) are not altered, so what about the ones before $1$? Say that $1$ is the $l$-th element of $S$ wrt the ordering of $\sigma$, call $k_{1},\ldots,k_{l-1}$ the position in $\sigma$ of the elements of $S$ before $1$, and call $k_{l}$ the position of $1$: the old score of all such elements ($1$ included) was
$$\frac{1}{1\cdot k_{1}}+\frac{1}{2\cdot k_{2}}+\ldots+\frac{1}{(l-1)\cdot k_{l-1}}+\frac{1}{l\cdot k_{l}},$$
and the new one is
$$\frac{1}{2\cdot k_{1}}+\frac{1}{3\cdot k_{2}}+\ldots+\frac{1}{l\cdot k_{l-1}}+\frac{1}{1\cdot 1}.$$
The score has increased by
$$1-\sum_{i=1}^{l-1}\frac{1}{i(i+1)k_{i}}-\frac{1}{lk_{l}},$$
and by our hypotheses $k_{l}>k_{l-1}>\ldots>k_{1}\geq 3$, so the worst increment happens for $k_{i}=i+2$ for all $i$; it's an elementary calculation then to show that $l=1$ is the worst case, for which the score increment becomes $\frac{2}{3}$.
If $S$ is not one of the first $n$ sets but it also contains $1$ and not $a$, then similarly its score increases, so there's no harm in ignoring them.
Finally, if $S$ contains $a$ but not $1$ (and there are at most $\frac{2}{3}n$ such $S$), after the swap all the elements of $S$ except $a$ have their score increased or unchanged, so we can ignore them again. The score of $a$ in a single $S$ before the swap was $1$, and after the swap is still positive, so the decrement is at most some $c<1$ (the same $c$ for all sets: we could bound $c$ away from $1$ in some way, if we felt like doing so, but I don't).
Putting all things together, the score increment is at least $\frac{2}{3}\cdot n-c\cdot\frac{2}{3}n>0$.
We can for sure do better with more care, but I'll leave it to more patient people. Also, of course "the first $\frac{4}{3}n$ sets" has no particular significance, any set of $\frac{4}{3}n$ sets including the $n$ sets where $1$ resides would do just fine.
