$ \newcommand{\R}{\mathbb{R}} \newcommand{\la}{\lambda} \newcommand{\p}{\partial} \newcommand{\PP}{\mathcal{P}}$ Let $x:=(x_1,\dots,x_n)\in[0,1]^n$, $y:=(y_1,\dots,y_n)\in[0,1]^n$, $h:=(h_1,\dots,h_n)$, \begin{equation*} h_i:=H(x_i,y_i),\quad H(u,v):=\frac2{\frac1u+\frac1v}=\frac{2uv}{u+v} \end{equation*} for $u>0$ and $v>0$, and, by continuity, $H(u,v):=0$ for $u\ge0$ and $v\ge0$ with $u v=0$. Let $Az:=\frac1n\sum_1^n z_i$ for $z:=(z_1,\dots,z_n)$. Then the result in question can written as \begin{equation*} L:=L(x,y):=H(Ax,Ay)-Ah\le L^*_n:= \left\{ \begin{alignedat}{2} &\frac12&&\text{ if $n$ is even}\\ & \frac12-\frac1{2n^2}&&\text{ if $n$ is odd}, \end{alignedat} \right. \tag{0} \end{equation*} with equality for some $x,y$ in $[0,1]^n$.
The maximum of $L(x,y)$ over all $(x,y)\in[0,1]^n\times[0,1]^n$ is attained. In what follows, let $(x,y)$ be such a maximizer.
With $[n]:=\{1,\dots,n\}$, $p$ and $q$ in $\{0,1\}$, and $|K|:=(\text{cardinality of $K$)}$, let
\begin{gather*}
I:=\{i\in[n]\colon 0<x_i<1\},\quad J:=\{i\in[n]\colon 0<y_i<1\},\\
I_p:=\{i\in[n]\colon x_i=p\},\quad J_q:=\{i\in[n]\colon y_i=q\},\\
s_{pq}:=\tfrac1n|I_p\cap J_q|,
\end{gather*}
so that $s_{00}+s_{01}+s_{10}+s_{11}\le1$.
If $Ax=0$, then $x=0$ and hence $h=0$ and $L=0$, which makes the inequality in (0) trivial. So, without loss of generality (wlog), $Ax>0$. Similarly, wlog $Ay>0$. So, \begin{equation*} r:=Ay/Ax\in(0,\infty). \tag{1} \end{equation*}
Let $\p_u$ denote the partial derivative with respect to a variable $u$.
Then
\begin{equation*}
\p_u H(u,v)=2\Big(\frac v{u+v}\Big)^2
\end{equation*}
for $u>0$ and $v>0$.
So, for any $i\in I$
\begin{equation*}
\frac n2\,\p_{x_i}L
=\Big(\frac r{r+1}\Big)^2-\Big(\frac{y_i}{x_i+y_i}\Big)^2=0,
\end{equation*}
because $(x,y)$ is a maximizer of $L$. So, $y=rx>0$ on $I$. Similarly, $y=rx>0$ on $J$, and hence $y=rx>0$ on $I\cup J$.
So, with $\xi:=\sum_{i\in I\cup J}x_i$$\xi:=\frac1n\,\sum_{i\in I\cup J}x_i$,
\begin{alignat*}{5}
&Ax=&& &&s_{10}&+&s_{11}&&+\xi, \tag{Ax}\\
&Ay=&&s_{01}&& &+&s_{11}&&+\xi r, \tag{Ay}\\
&Ah=&& && &&s_{11}&&+\xi\frac{2r}{1+r}.
\end{alignat*}
So,
\begin{align*}
L&=\frac{2 Ax\,Ay}{Ax+Ay}-Ah \\
&=Ax\frac{2r}{1+r}-\Big(s_{11}+\xi\frac{2r}{1+r}\Big) \\
&=(s_{10}+s_{11})\frac{2r}{1+r}-s_{11}. \tag{2}
\end{align*}
It also follows from (Ax) and (Ay) that the equality in (1) can be rewritten as
\begin{equation*}
s_{01}+s_{11}=r(s_{10}+s_{11}).
\end{equation*}
So, if $s_{10}+s_{11}=0$, then $s_{11}=0$ and hence, by (2), $L=0$. So, wlog $s_{10}+s_{11}>0$ and hence
$r=\frac{s_{01}+s_{11}}{s_{10}+s_{11}}$. Using this expression for $r$, we get from (2):
\begin{align*}
L=M:=\frac{2 s_{01} s_{10} + (s_{01}+ s_{10})s_{11}}{s_{01} + s_{10} + 2 s_{11}}.
\end{align*}
Next,
\begin{equation*}
\p_{s_{11}}M:=\frac{(s_{01}-s_{10})^2}{(s_{01} + s_{10} + 2 s_{11})^2}\ge0.
\end{equation*}
So, wlog one may replace $s_{11}$ by its largest possible value, $1-s_{01}-s_{10}$:
\begin{equation*}
L=M\le N:=M|_{s_{11}=1-s_{01}-s_{10}}=
\frac{(1-s_{01})s_{01}+(1-s_{10})s_{10}}{2-s_{01}- s_{10}}.
\end{equation*}
Further,
\begin{equation*}
(\p_{s_{01}}+\p_{s_{10}})N=
\frac{4(1-s_{01})(1-s_{10})}{(2-s_{01}-s_{10})^2}\ge0.
\end{equation*}
So, if we increase $s_{01}$ and $s_{10}$ by the same amount, while keeping $s_{01}+s_{10}\le1$, the value of $N$ may only increase. So,
\begin{equation*}
L\le N|_{s_{10}=1-s_{01}}=2(1-s_{10})s_{10}
\le2(1-\tfrac mn)\tfrac mn=L^*_n,
\end{equation*}
where $m:=\lfloor n/2\rfloor$; the latter inequality follows because $(1-u)u$ is decreasing in $|u-1/2|$ for $u\in[0,1]$.
The inequality in (0) turns into the equality if $(x_i,y_i)=(1,0)$ for $i=1,\dots,m$ and $(x_i,y_i)=(0,1)$ for $i=m+1,\dots,n$.
The entire proof is now complete.