Perfect reconstruction: what properties must the analysis LPF satisfy
So the central question here is what the restrictions for the filters of this class of filterbanks (i.e. $N=2$-filter bank with the HPF generated from a lowpass/highpass transform of the LPF) is.

Let us for that first introduce a couple of helping definitions: 

$$\begin{align}
\mathbf{h}(z) &amp;= \begin{bmatrix}h_0(z)\\h_1(z)\end{bmatrix}&amp;\text{analysis filterbank matrix}\\
\mathbf{r}(z) &amp;= \begin{bmatrix}r_0(z)\\r_1(z)\end{bmatrix}&amp;\text{reconstruction filterbank matrix}\\
\end{align}$$

With these in place: Let's consider the polyphase TypeI representation $A(z)$ (analysis) and TypeII representation $R(z)$ (reconstruction) of these filter banks (by the way, this is why we have these different types of polyphase deconstructions); I hope my notation is clear, $h_{1,0}$ is the zeroth polyphase component of the high pass filter.

$$\begin{align}
\mathbf{h}(z) &amp;= 
   \begin{bmatrix}
     h_{0,0}^{\mathrm I}(z^2) &amp; h_{0,1}^{\mathrm I}(z^2)\\
     h_{1,0}^{\mathrm I}(z^2) &amp; h_{1,1}^{\mathrm I}(z^2)
   \end{bmatrix}
   \begin{bmatrix}
     z^{-0}\\
     z^{-1}
   \end{bmatrix}\\
&amp;= \mathbf A(z^2) \mathbf e_z(z)\\
\mathbf{r}^T(z) &amp;=
   \begin{bmatrix}
     z^{-1} &amp; 
     z^{-0}
   \end{bmatrix}
   \begin{bmatrix}
     r_{0,0}^{\mathrm {II}}(z^2) &amp; r_{0,1}^{\mathrm {II}}(z^2)\\
     r_{1,0}^{\mathrm {II}}(z^2) &amp; r_{1,1}^{\mathrm {II}}(z^2)
   \end{bmatrix}\\
&amp;= \mathbf{ \tilde e}_z(z) \mathbf R(z^2) \\
\end{align}$$

Now, looking at this, we simply define perfect reconstruction "intuitively" as the case where applying $\mathbf R$ to the output of $\mathbf A$ leads to the input of $\mathbf A$, of course delayed, and maybe scaled with a constant $c$:

$$\begin{align}
  h,\,r &amp;\text{ form a perfect reconstruction fb}\\
  &amp;\iff\\
  \mathbf R(z)\mathbf A(z) &amp;= c z^{-k} \mathbf I_N &amp; c \text{ const.},\, k\in\mathbb N
\end{align}$$

That gives us a clear method to move forward:

$$\begin{align}
  \mathbf R(z)\mathbf A(z) &amp;=    \begin{bmatrix}
     h_{0,0}^{\mathrm {II}}(z) &amp; h_{0,1}^{\mathrm {II}}(z)\\
     h_{1,0}^{\mathrm {II}}(z) &amp; h_{1,1}^{\mathrm {II}}(z)
   \end{bmatrix}
   \begin{bmatrix}
     r_{0,0}^{\mathrm I}(z) &amp; r_{0,1}^{\mathrm I}(z)\\
     r_{1,0}^{\mathrm I}(z) &amp; r_{1,1}^{\mathrm I}(z)
   \end{bmatrix}\\
&amp;= \begin{bmatrix}
h_{0,0}^{\mathrm {II}}(z)r_{0,0}^{\mathrm I}(z) +  h_{0,1}^{\mathrm {II}}(z)r_{1,0}^{\mathrm I}(z) &amp;
h_{0,0}^{\mathrm {II}}(z)r_{0,1}^{\mathrm I}(z) + 
h_{0,1}^{\mathrm {II}}(z)r_{1,1}^{\mathrm I}(z)\\
h_{1,0}^{\mathrm {II}}(z)r_{0,0}^{\mathrm I}(z) +
h_{1,1}^{\mathrm {II}}(z)r_{1,0}^{\mathrm I}(z) &amp;
h_{1,0}^{\mathrm {II}}(z)r_{0,1}^{\mathrm I}(z) +
h_{1,1}^{\mathrm {II}}(z)r_{1,1}^{\mathrm I}(z)
\end{bmatrix}\\
&amp;\overset!= \begin{bmatrix}c z^{-k}&amp;0\\0&amp;c z^{-k}\end{bmatrix}
\end{align}$$

EDIT now, let's do our jobs of inserting our conditions:

$$r_0  = h_0 (L-1-n)$$

notice that the substitution of $n$ by $L-1-n$ implies that the typeI polyphase decomposition of $r_0$ is the same as the typeII decomposition of $h_0$! 

Hence, the diagonal entries of above matrix get a lot easier to calculate:

$$\begin{align}
  \mathbf R(z)\mathbf A(z) &amp;=\begin{bmatrix}
h_{0,0}^{\mathrm {II}}(z)r_{0,0}^{\mathrm I}(z) +  h_{0,1}^{\mathrm {II}}(z)r_{1,0}^{\mathrm I}(z) &amp;
h_{0,0}^{\mathrm {II}}(z)r_{0,1}^{\mathrm I}(z) + 
h_{0,1}^{\mathrm {II}}(z)r_{1,1}^{\mathrm I}(z)\\
h_{1,0}^{\mathrm {II}}(z)r_{0,0}^{\mathrm I}(z) +
h_{1,1}^{\mathrm {II}}(z)r_{1,0}^{\mathrm I}(z) &amp;
h_{1,0}^{\mathrm {II}}(z)r_{0,1}^{\mathrm I}(z) +
h_{1,1}^{\mathrm {II}}(z)r_{1,1}^{\mathrm I}(z)
\end{bmatrix}\\
&amp;=\begin{bmatrix}
h_{0,0}^{\mathrm {II}}(z)h_{0,0}^{\mathrm {II}}(z) +
h_{0,1}^{\mathrm {II}}(z)-h_{1,0}^{\mathrm{II}}(z) &amp;
h_{0,0}^{\mathrm {II}}(z)h_{0,1}^{\mathrm {II}} + 
h_{0,1}^{\mathrm {II}}(z)-h_{1,1}^{\mathrm{II}}(z)\\
h_{1,0}^{\mathrm {II}}(z)h_{0,0}^{\mathrm{II}}(z) +
h_{1,1}^{\mathrm {II}}(z)-h_{1,0}^{\mathrm{II}}(z) &amp;
h_{1,0}^{\mathrm {II}}(z)h_{0,1}^{\mathrm{II}}(z) +
h_{1,1}^{\mathrm {II}}(z)-h_{1,1}^{\mathrm{II}}(z)
\end{bmatrix}\\
&amp;=\begin{bmatrix}
h_{0,0}^{\mathrm {II}}(z)h_{0,0}^{\mathrm {II}}(z) -
h_{0,1}^{\mathrm {II}}(z)h_{1,0}^{\mathrm{II}}(z) &amp;
h_{0,0}^{\mathrm {II}}(z)h_{0,1}^{\mathrm {II}} -
h_{0,1}^{\mathrm {II}}(z)h_{1,1}^{\mathrm{II}}(z)\\
h_{1,0}^{\mathrm {II}}(z)h_{0,0}^{\mathrm{II}}(z) -
h_{1,1}^{\mathrm {II}}(z)h_{1,0}^{\mathrm{II}}(z) &amp;
h_{1,0}^{\mathrm {II}}(z)h_{0,1}^{\mathrm{II}}(z) 
h_{1,1}^{\mathrm {II}}(z)h_{1,1}^{\mathrm{II}}(z)
\end{bmatrix}\\
\end{align}$$

To achieve the zero entries in the identity matrix, it is necessary that

$$\begin{align}
h_{0,1}^{\mathrm {II}} h_{0,0}^{\mathrm {II}}(z)&amp;= h_{0,1}^{\mathrm {II}}(z)h_{1,1}^{\mathrm{II}}(z)\\
h_{1,0}^{\mathrm {II}}(z)h_{0,0}^{\mathrm{II}}(z) &amp;= h_{1,0}^{\mathrm{II}}(z)h_{1,1}^{\mathrm {II}}(z)
\end{align}$$

This breaks down to filters that need to have zeros at every $N$th position, since the 2-polyphased decomposition of a HPF/LPF transformed HPF is either the same as the original's decomposition, or the original's decomposition with a factor of $-1$.