= Solution
For an inverted <free surface>, <gravity> promotes a displacement instead of restoring it. The <surface tension> term still resists curvature. Hence the <gravity-capillary wave> relation becomes, with $q=|k|$,
$$
\boxed{\omega^2=q\left(\frac{\sigma q^2}{\rho}-g\right).}
$$
For $0<q<k_c$, where $k_c=\sqrt{\rho g/\sigma}$, $\omega^2<0$. Writing the displacement as $\eta\propto e^{st+ikx}$ gives the <Rayleigh-Taylor instability> growth rate
$$
s=\sqrt{gq-\frac{\sigma q^3}{\rho}},\qquad 0<q<k_c.
$$
Shorter <wavelengths>, $q>k_c$, remain oscillatory because <surface tension> dominates. The marginal scale is determined by the <capillary length>,
$$
\ell_c=\sqrt{\frac{\sigma}{\rho g}}
=\sqrt{\frac{7.2\times10^{-2}}{1000(9.81)}}\ \mathrm m
\simeq2.71\ \mathrm{mm},
\qquad
\boxed{\lambda_c=2\pi\ell_c\simeq1.70\ \mathrm{cm}.}
$$
A sufficiently small aperture excludes the long unstable modes: this is <finite-aperture stabilization of an inverted liquid surface>. The estimate requested is therefore \b[a container opening of order one centimetre], with $1.7\ \mathrm{cm}$ the marginal full wavelength rather than a geometry-independent maximum diameter.
The exact largest opening cannot be fixed from the stated infinite-surface <dispersion relation> alone. It requires the aperture shape and contact-line <boundary conditions>. For example, a one-dimensional span of width $L$ whose <contact line> is pinned at both ends has a half-sine lowest mode with $q_{\min}=\pi/L$. In that simple model,
$$
\boxed{L_{\rm crit}=\pi\ell_c\simeq8.5\ \mathrm{mm},\qquad
L<L_{\rm crit}\ \text{excludes growing modes}.}
$$
Different geometries change the numerical factor. This is a small-disturbance stability estimate with negligible air inertia and no leakage or contact-line failure; a guarantee that a real inverted container retains its water needs those additional boundary details.
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