Yes, conventions in the cited paper are such that their $k$ is twice as big as the $k$ from the question.

Yes. I'm sorry, too many mistakes in formulation.

I have corrected the formulation.

Уеs, of course. In fact it was shown numerically in arxiv.org/abs/2407.10787 that such a solution exists for example for $k\approx 29.056$. But how this can be proved analytically?

The coefficient $2\pi$ is missing in the above formula.

More convenient form: $$I(n,m)=\sum_{1+\lceil \frac{m(n-1)}{2n}\rceil}^{1+\lfloor \frac{m(n+1)}{2n}\rfloor}a^{m+2}_sC^m_{n(s-1)-\frac{m(n-1)}{2}}\cos{[(m+2-2s)n\theta]},$$ where $$C^n_k=\frac{1}{4^n}\binom{2k}{k}\binom{2(n-k)}{n-k},$$ and $a^n_k$ is given in the comments.

If $\lceil \frac{m(n-1)}{2n}\rceil > \lfloor \frac{m(n+1)}{2n}\rfloor$, the integral is zero.

I got (and verified numerically) a result that looks a little different: $$2\pi \sum_{s=1+\lceil\frac{m(n-1)}{2n}\rceil}^{1+\lfloor\frac{m(n+1)}{2n}\rfloor} a^{m+2}_s \cos(n(m+2-2s)\theta)\left (\sum_{l=0}^{\lfloor m/2\rfloor}b^m_la^{m-2l}_{n(s-1)-m(n-1)/2-l}\right).$$ Here $$a^n_m=2^{-n}\binom{n}{m},\;\;b^m_l=2^{-m}(-1)^l\binom{m}{l}\binom{2m-2l}{m}.$$ The result is valid if $m(n-1)$ is even. If $m(n-1)$ is odd, the integral is zero.