Evaluating $\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}$ I ma trying to prove

$$S=\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}=\frac1{24}\ln^42-\frac14\ln^22\zeta(2)+\frac{21}{8}\ln2\zeta(3)-\frac{9}{8}\zeta(4)+\operatorname{Li}_4\left(\frac12\right)$$

where $\overline{H}_n$ is the alternating harmonic number and $H_n$ is the harmonic number.
I need this sum to complete my solution here.
Here is my trial,
Following @user97357329's note in the comments of the same link above 
$$\sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty f(2n-1)+\sum_{n=1}^\infty f(2n)$$
Giving us 
$$S=\underbrace{\sum_{n=1}^\infty\frac{\overline{H}_{2n-1}H_{n-1/2}}{(2n-1)^2}}_{\large S_1}+\frac14\underbrace{\sum_{n=1}^\infty\frac{\overline{H}_{2n}H_{n}}{n^2}}_{\large S_2}$$
I managed to evaluate $S_2$ using $\overline{H}_{2n}=H_{2n}-H_n$.
Regarding $S_1$, I used $\overline{H}_{2n-1}=H_{2n}-H_n+\frac1{2n}$ and $H_{n-1/2}=2H_{2n}-H_n-2\ln2$ 
therefore
$$S_1=2\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n-1)^2}-\color{blue}{\sum_{n=1}^\infty\frac{H_nH_{2n}}{(2n-1)^2}}-2\ln2\sum_{n=1}^\infty\frac{H_{2n}}{(2n-1)^2}+\color{red}{\sum_{n=1}^\infty\frac{2H_{2n}-H_n-2\ln2}{2n(2n-1)^2}}$$
and I am stuck with the blue and red sums, any idea? Thank you.
 A: The blue sum, with the denominator rearranged, comes immediately from the result given in Section 4.59, page $313$, from the book (Almost) Impossible Integrals, Sums, and Series.

$$\zeta(4)$$
$$=\frac{8}{5}\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^2}+\frac{64}{5}\sum _{n=1}^{\infty } \frac{ \left(H_{2 n}\right)^2}{ (2 n+1)^2}+\frac{64}{5}\sum _{n=1}^{\infty } \frac{H_{2 n}}{(2 n+1)^3}$$
$$-\frac{8}{5}\sum _{n=1}^{ \infty } \frac{\left(H_{2 n}\right){}^2}{ n^2}-\frac{32}{5}\underbrace{\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{(2 n+1)^2}}_{\text{The series you need}}-\frac{64}{5}\log(2)\sum _{n=1}^{ \infty } \frac{H_{2 n}}{(2 n+1)^2}-\frac{8}{5}\sum _{n=1}^{\infty } \frac{H_{2 n}^{(2)}}{n^2}.$$

In fact, in the book the author nicely exploits the fact that for the linear Euler sum of the type $\displaystyle \sum_{n=1}^{\infty} \frac{H_n}{n^m}$, with $m=3$, we arrive at $5/4 \zeta(4)$ which allows us to express the $\zeta(4)$ value in terms of a sum of seven series. You might not need this precise representation, but almost all the steps presented in the solution. It is precisely the same strategy as for the weight $5$ case which is given in On the calculation of two essential harmonic series with a weight 5 structure, involving harmonic numbers of the type $H_{2n}$. In this case we play with weight $4$ series. Observe that all the other series above are known or easily reducible to known series. 
A note: In this question Two very advanced harmonic series of weight $5$, if you take a look at the second and third series you may see how they look like when having $2n-1$ and $2n+1$ in denominator (the latter version looks way better in terms of a closed-form). Well, like our case, except that there we are on the realm of weight $5$ series.

What about the red part? We want a clever rearrangement of the initial series, that is 
$$\sum _{n=1}^{\infty } \frac{2 H_{2 n}-H_n-2 \log (2)}{2 n (2 n-1)^2}$$
$$=2\sum _{n=1}^{\infty } \frac{H_{2 n-1}+1/(2n)}{(2 n-1)^2}-\sum _{n=1}^{\infty } \frac{H_n}{(2 n-1)^2}-\sum _{n=1}^{\infty } \frac{H_n}{2 n (2 n-1)}-2 \log (2)\sum _{n=1}^{\infty } \frac{1}{(2 n-1)^2}$$
$$+2 \sum _{n=1}^{\infty } \frac{H_n-H_{2 n}+\log (2)}{2 n (2 n-1)}.$$
Both the first and the second series are done by using the results from this paper A new powerful strategy of calculating a classof alternating Euler sums by Cornel Ioan Valean, particularly the main theorem and lemma $4$. Then, the third and the fourth sums are trivial. 
Finally, there is a nice thing to observe about the fifth sum, that is if we reindex it and start from $n=0$, we can simply use the series from the second step in this answer Prove $\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\frac{\pi^2}{24}$, which is finalized elementarily.
End of story.
