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][1]. 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. 

[1]: https://www.math.temple.edu/~tewodros/Harmonic_Sumsvnew.pdf