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.

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