It seems that the following theorem can be proved without using any information about primitive or imprimitive subgroups.
THEOREM. Let $H \subseteq S_N$ be a transitive subgroup, and assume that $N \triangleleft H$ and $H/N \cong S_{n/2}$, where $N/2 \le n < N$. Then $N$ is an elementary abelian $2$-group, and if $N$ is nontrivial, then $n = N/2$.
Proof. Let $S$ be an arbitrary point stabilizer in $S_N$ so $|S_N:S| = N$. Now $|H:(H \cap S)| = N$ since $H$ is transitive. Write $d = |N:(N \cap S)|$, so $d = |N(H \cap S):(H \cap S)|$, and this divides $|H:(H \cap S)| = N$. Also, $N/d = |H:N(H \cap S)|$ is the index of a subgroup of $H/N \cong S_n$, so this index cannot be smaller than $n$. Thus $N/d \ge n \ge N/2$, and thus $d \le 2$.
Since $S$ was an arbitrary point stabilizer in $S_N$ and $|N:(N \cap S)| = d \le 2$, we see that all orbits of $N$ have size $1$ or $2$, and thus $N$ is an elementary abelian $2$-group. If $N$ is nontrivial, then $d = 2$ for some choice of point stabilizer $S$, and it follows that $n = N/2$. QED
The case where $N$ is trivial occurs with $N = 6$ and $n = 5$, but I doubt that there are any other possibilities, though I don't see a proof. If $S_N$ has a transitive subgroup isomorphic to $S_n$, this would imply that $S_n$ has a subgroup of index $N$, where $n < N \le 2n$. I think that forces $N = n(n-1)/2$ except when $n = 5$ and $N = 6$.