This is gonna be a few steps.
Current discrete time is $n$, hop is $H$, length of window is $L$. fraction of overlap is $\tfrac{L-H}{L}$
if we make
$$ \sum\limits_{m=-\infty}^{+\infty} w\left( \tfrac{n-mH}{L} \right) = 1 $$
then
$$\begin{align}
x[n] &= x[n] \times 1 \\
&= x[n] \times \sum\limits_{m=-\infty}^{+\infty} w\left( \tfrac{n-mH}{L} \right) \\
&= \sum\limits_{m=-\infty}^{+\infty} x[n] w\left( \tfrac{n-mH}{L} \right) \\
&= \sum\limits_{m=-\infty}^{+\infty} x_m[n-mH] \\
\end{align}$$
where
$$\begin{align}
x_m[n-mH] & \triangleq x[n] w\left( \tfrac{n-mH}{L} \right) \\
x_m[n] & = x[n+mH] w\left( \tfrac{n}{L} \right) \\
\end{align}$$
So $x_m[n]$ is a finite-length frame of audio of length $L$ and represents the audio in the neighborhood of $x[n+mH]$ or $m$ hops from the beginning.
Each input frame $x_m[n]$ gets processed into an output frame $y_m[n]$ and the output frames overlap-add to reconstruct the output in the same manner as the input is constructed.
$$ y[n] = \sum\limits_{m=-\infty}^{+\infty} y_m[n-mH] $$
So how do you make $ \sum\limits_{m=-\infty}^{+\infty} w\left( \tfrac{n-mH}{L} \right) = 1 $ ?
(this isn't done yet. i'll get back to this tomorrow.)