A solution by Cornel Ioan Valean
If we set $a_n=H_n/n$ and $b_n=\zeta(2)-H_{2n}^{(2)}$, and then apply Abel's summation, we arrive at
$$\sum_{n=1}^{\infty}\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)$$
$$=\frac{1}{2}\underbrace{\sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{(2n+1)^2}}_{\displaystyle A}+\underbrace{\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{(2n+1)^2}}_{\displaystyle B}+\frac{1}{8}\underbrace{\sum_{n=1}^{\infty}\frac{H_n^2+H_n^{(2)}}{n^2}}_{\displaystyle C}-\frac{1}{4}\underbrace{\sum_{n=1}^{\infty}\frac{H_n}{n^3}}_{\displaystyle 5/16\zeta(3)}.\tag1$$
For the sum $A$, we consider a result from the book (Almost) Impossible Integrals, Sums, and Series, page $291$,
\begin{equation*}
A=\sum_{k=1}^{\infty} \frac{H_k^2-H_k^{(2)}}{(k+1)(k+n+1)}
\end{equation*}
\begin{equation*}
=\frac{(\psi(n+1)+\gamma)^3+3(\psi(n+1)+\gamma)(\zeta(2)-\psi^{(1)}(n+1))+2(\zeta(3)+1/2\psi^{(2)}(n+1))}{3n},
\end{equation*}
where if we multiply both sides by $n$, differentiate once with respect to $n$, and set $n=-1/2$, we arrive at
\begin{equation*}
\sum_{n=1}^{\infty} \frac{H_n^2-H_n^{(2)}}{(2n+1)^2}=\frac{15}{4}\zeta (4)-7 \log(2) \zeta(3)+3\log^2(2)\zeta(2).
\end{equation*}
For the sum $B$, we apply Abel's summation with $a_n=1/(2n+1)^2$ and $H_n^{(2)}$ that gives
\begin{equation*}
B=\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{(2n+1)^2}=\frac{15}{8}\zeta(4)-\frac{7}{4}\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n^2}+2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{H_n^{(2)}}{n^2}
\end{equation*}
\begin{equation*}
=\frac{1}{3}\log^4(2)-2\log^2(2)\zeta(2)+7\log(2)\zeta(3)-\frac{121}{16}\zeta(4)+8 \operatorname{Li}_4\left(\frac{1}{2}\right),
\end{equation*}
where
\begin{equation*}
\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_n^{(2)}}{n^2}=\frac{1}{6}\log^4(2)-\log^2(2)\zeta(2)+\frac{7}{2}\log(2)\zeta(3)-\frac{51}{16}\zeta(4)+4\operatorname{Li}_4\left(\frac{1}{2}\right),
\end{equation*}
may be found calculated in (Almost) Impossible Integrals, Sums, and Series, pages $505$-$506$.
Finally, for the sum $C$ we may elegantly exploit the fact that $\displaystyle\int_0^1 x^{n-1}\log^2(1-x)\textrm{d}x =\frac{H_n^2+H_n^{(2)}}{n}$, you may also find nicely calculated in (Almost) Impossible Integrals, Sums, and Series, page $60$. In general, such integrals may also be viewed in terms of Beta function and hence a solution is pretty easily extracted. Then,
$$C=\sum_{n=1}^{\infty}\frac{H_n^2+H_n^{(2)}}{n^2}=-\int_0^1 \frac{\log^3(1-x)}{x}\textrm{d}x=-\int_0^1 \frac{\log^3(x)}{1-x}\textrm{d}x=6\zeta(4). $$
Collecting all sums in $A$, $B$, $C$ and plugging them in $(1)$, we arrive at the desired result
$$\sum_{n=1}^\infty\frac{H_n}{n}\left(\zeta(2)-H_{2n}^{(2)}\right)=\frac{1}{3}\log^4(2)-\frac{1}{2}\log^2(2)\zeta(2)+\frac{7}{2}\log(2)\zeta(3)-\frac{21}{4}\zeta(4)+8\operatorname{Li}_4\left(\frac{1}{2}\right).$$
COOL BONUS
Combining the sums $A$ and $B$ we obtain that
$$\sum_{n=1}^{\infty} \frac{H_n^2}{(2n+1)^2}=\frac{1}{3}\log^4(2)+\log^2(2)\zeta(2)-\frac{61}{16}\zeta(4)+8 \operatorname{Li}_4\left(\frac{1}{2}\right).$$
