This problem is relevant to the spin operator matrix elements in the quantum 1D XY model.
For any even integer $N$, define two sets $$K_+=\left\{\frac{(2m+1)\pi}{N}|m=-\frac{N}{2},-\frac{N}{2}+1,...,\frac{N}{2}-1\right\}$$ and $$K_-=\left\{\frac{2m\pi}{N}|m=-\frac{N}{2},-\frac{N}{2}+1,...,\frac{N}{2}-1\right\}.$$ For any integer $1\leq n\leq (N-1)$, if $n$=even (odd), we choose $n$ distinct numbers $k_1,k_2,...,k_n$ from set $K_+$ ($K_-$), and choose $n+1$ distinct numbers $k'_1,k'_2,...,k'_{n+1}$ from set $K_-$ ($K_+$). Prove that: \begin{align}\sum_{1\leq j_1<j_2<...<j_{n+1}\leq N}\det\left( \begin{array}{ccc} e^{-ik'_1j_1} & ... & e^{-ik'_1j_{n+1}} \\ .. & .. & .. \\ e^{-ik'_{n+1}j_1} & ... & e^{-ik'_{n+1}j_{n+1}}\\ \end{array} \right)\det \left( \begin{array}{cccc} 1 & e^{ik_1j_1} & ... & e^{ik_nj_1} \\ -1 & e^{ik_1j_2} & ... & e^{ik_nj_2} \\ .. & .. & .. & .. \\ (-1)^n & e^{ik_1j_{n+1}} & .. & e^{ik_nj_{n+1}} \\ \end{array} \right)=2^nN\delta\left(\sum^{n+1}_{j=1} k'_j,\sum^n_{i=1} k_i\right)\frac{\prod_{i>i'}(e^{ik_i}-e^{ik_{i'}})\prod_{j>j'}(e^{-ik'_j}-e^{-ik'_{j'}})}{\prod^n_{i=1}\prod^{n+1}_{j=1}(1-e^{i(k'_j-k_i)})}.\end{align}
The Cauchy determinant and Cauchy–Binet formula seem highly relevant, but I have no idea how to treat the sum '$\sum_{j_1<j_2<...<j_{n+1}}$', i.e., how to split it into a summation over $n$ variables '$\sum_{j_1<j_2<...<j_{n}}$' (leading to the Cauchy determinant) and a single sum '$\sum^N_{l=1}$' (leading to the $\delta$-function).
-Update: This problem has recently been solve in the following paper by using a fermionic approach: https://arxiv.org/abs/1709.00682