How to prove $\sum_{k=1}^{\frac{p-1}{2}}\frac{(-1)^k}{k}\sum_{i=\lfloor k/2\rfloor +1}^k\frac{1}{2i-1}\equiv 0\pmod{p}$?

It turned out the OP's question required a good amount of work and a joint effort (with Roberto Tauraso). However, "the margin here is too small to contain the proof", so to say.

Instead, readers can find the resolution at this link. In addition, we prove that \begin{align*} \sum_{k=1}^{\infty}\frac{(-1)^k}{k}\sum_{i=\lfloor k/2\rfloor +1}^k\frac{1}{2i-1}=-\frac58\zeta(2). \end{align*}

Comments and suggestions are welcome.