How can there be more than one stable isobar for a given atomic mass number? It can't simply be because the energy gap is less than the mass-energy of an electron. Electrons can participate in decay through electron capture. And there's always one least energetic combination of protons, neutrons, and electrons. My initial guess was that it was because stable isobars always differ in atomic number by at least 2, so the one in between them must have more  state than both, so it would take two simultaneous electron captures or $\beta^\pm$ emissions to go from one stable isobar to the other, which would not be reasonably possible. But it turns out that can happen too, but it takes a very, very long time. So what's going on?
 A: Let’s look at the mass-36 system, with data from the Nuclear Wallet Cards:

The column $\Delta$ is the “mass excess.” They are tabulated for neutral atom, which means you can compute the $Q$-values for electron capture and $\beta^-$ decay just by subtracting the mass excesses.  For $\beta^+$ decay, the $Q$-value is reduced by 1 MeV to account for the mass of the extra spectator electron and the emitted positron in the final state. So for this system, the decays of interest are
\begin{align}
\rm^{36}Cl &\overset{\beta^-}\longrightarrow{} \rm^{36}Ar + 0.7\, MeV
\\
\rm^{36}Cl &\overset{\mathrm{e.c.}}\longrightarrow{} \rm^{36}S + 1.1\, MeV
\\
\rm^{36}Cl &\overset{\beta^+}\longrightarrow{} \rm^{36}S + 0.1\, MeV
\end{align}
Your question is whether you can permit
$$
\rm^{36}Ar \overset{?}{\longrightarrow}{} ^{36}S + 0.4\,MeV
$$
Nothing seems to prohibit this decay. The problem lies in observing it.  It is energetically allowed, by double electron capture.  But the $Q$-value is quite small, and the bulk of the energy would be carried away by the two electron neutrinos.  A ballpark lifetime for double-weak nuclear decays is $10^{21}$ years — that is, with a mole of the nuclide, you would expect one decay per day or less.  The only detectable prompt radiation would be soft x-rays as the vacancies in the sulfur’s electron cloud were filled. Furthermore, a quick look suggests the electronic binding energies associated with those soft x-rays might be a meaningful reduction in the $Q$-value of the double-e.c. decay.  In general, the smaller the energy released in the decay, the longer the lifetime.
If you wanted to design an experiment to look for this particular decay, you would probably need an enriched argon volume: only 0.3% of natural argon is argon-36. Your decay volume would be either a low-density gas or a cryogenic liquid target.  Backgrounds would be problematic.  Cosmic rays which interact with Earth’s atmosphere or surface can produce neutrons by spallation.  These neutrons thermalize and then wander around until they capture on something. Argon-36 has a pretty healthy cross section for neutron capture to argon-37.  But argon-37 is unstable and decays, with a lifetime of about a month, via electron capture to chlorine-37! This cosmic-induced background decay probably produces a similar x-ray spectrum to the putative double-e.c. to sulfur.  So now you are looking at

*

*isotopically enriched argon

*liquified in a cryostat

*surrounded by x-ray detectors

*probably looking for x-ray coincidences, where one atom has two vacancies at once

*perhaps also looking at x-ray energy spectra, to distinguish chlorine x-rays from sulfur x-rays

*except that Compton scattering means that all gamma-ray spectra have a continuum component, so low-rate spectroscopy sucks

*at the bottom of a mineshaft to escape cosmic rays

*surrounded by neutron absorbers because escape from cosmic rays is impossible

*to produce 100–1000 events per year per mole

This experiment would be hard, and probably no one has done it, because there are more promising double-beta candidates. This 2017 review (pdf link) clarifies that they use “stable” to mean that no finite half-life has been found, and for argon-36 they qualify their “stable” with the decay mode $2\beta^+=?$.
