Let's assume continuous time (rather than discrete time).
If you do not process the windowed data at all, you would like the output (the sum of the windowed frames) to be equal to the original signal. Allowing scaling of the output by a constant scaling term, this is only possible if the sum of all of the time-shifted window functions is constant over time.
A typical window function (including Hann) is a sum of harmonic cosine terms. Let's assume you only use such windows. At the end of the window, the cosine terms have derivative discontinuities that can only be "cured" by stitching to the end of the window the start of another window which has identical values of all derivatives. This, and that you probably want the windows to be distributed evenly, requires that the hopsize is $\frac{1}{N}$ window lengths, with $N$ integer. The truncated cosine terms in each of the first $N$ windows are extended to complete cosines by concatenation of every $N\text{th}$ window.
The sum $f(T)$ of the windows distributed this way is:
$$f(T) = \sum_{n = 0}^{N - 1}\sum_{h = 0}^{H - 1} a_h\cos\left(2\pi h\left(T-\frac{n}{N}\right)\right),$$
where $h$ is the harmonic number, $H$ is the number of harmonics in the window function, $a_h$ are the window function specific coefficients, $T$ is time expressed so that a time difference of $1$ means one window length. $h$ and $n$ are the running harmonic and hop numbers.
Let's try window densities $N = 1\dots3$ with a window that has two harmonics ($H=2$). First $N=1:$
$$f(T) = a_0 + a_1\cos(2\pi T)$$
then $N=2:$
$$f(T) = a_0 + a_1\cos(2\pi T) + a_0 + a_1\cos\left(2\pi \left(T - \frac{1}{2}\right)\right) = 2a_0$$
and $N=3:$
$$f(T) = a_0 + a_1\cos(2\pi T) + a_0 + a_1\cos\left(2\pi \left(T - \frac{1}{3}\right)\right) + a_0 + a_1\cos\left(2\pi \left(T - \frac{2}{3}\right)\right) = 3a_0$$
This could be for example the Hann window with $a_0 = \frac{1}{2}$ and $a_1 = \frac{1}{2}:$
Figure 1. Hann windows sum to constant with a hopsize of 1/2 or 1/3 window lengths, but not with a hopsize of 1 window length.
Although it is not as evident as with the 1/2 hopsize, the $1\text{st}$ harmonic cosines cancel each other completely with the 1/3 hopsize. The cancellation can be illustrated by representing the phases and amplitudes of the harmonic terms as complex numbers:
Figure 2. Complex number representation of the amplitudes and phases of the 1st harmonic terms in shifted concatenated Hann windows, with a hopsize of $1/3$ window lengths. The rotational symmetry of the pattern means that the first harmonics cancel, because the sum of the three complex numbers must be zero.
For window functions with $H$ harmonic cosine terms, if we tabulate for different window densities $N$ the factors appearing in front of each $h\text{th}$ harmonic cosine term in $f(T),$ a pattern emerges:
$$\begin{array}{r|lllllllllll}&\rlap{h}\\
N&0&1&2&3&4&5&6&7&8&9\\
\hline
1&1&1&1&1&1&1&1&1&1&1\\
2&2&0&2&0&2&0&2&0&2&0\\
3&3&0&0&3&0&0&3&0&0&3\\
4&4&0&0&0&4&0&0&0&4&0\\
5&5&0&0&0&0&5&0&0&0&0\\
6&6&0&0&0&0&0&6&0&0&0\\
7&7&0&0&0&0&0&0&7&0&0\\
8&8&0&0&0&0&0&0&0&8&0\\
9&9&0&0&0&0&0&0&0&0&9\end{array}$$
The factors in the 2-term window example appear in the two first columns and three first rows.
For $H \le N,$ all harmonics except for the $0\text{th}$ harmonic cancel completely. This means for example that the 4-cosine-term ($H = 4$) Blackman–Harris window could be used with a 1/4 hopsize or greater ($N \ge 4$), requiring only an output scaling by $\frac{1}{4a_0} = \frac{1}{4\times0.35876}.$
