82 and 126 are magic numbers for nuclei, so $^{208}\mathrm{Pb}$ is doubly stable. So it seems to me like the hypernucleus $^{209}_{\Lambda^0}\mathrm{Pb}$ would be stable as well, because the $\Lambda^0$ baryon would sit at the bottom of its own shell, and its decay would have to push the resulting proton or neutron all the way to the top of the respective energy shell. The decay energy of a lambda baryon is only 177 MeV, and I'm not sure if that's enough.
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We can do a quick and dirty estimate of this by treating the lambda as just another baryon and using the Nilsson model, in which the spacing of the major shells is $\hbar\omega_0\approx(42\ \text{MeV})A^{-1/3}$. This comes out in this case to be about 7 MeV. The Fermi level in lead is probably around $N=5$ ($N$ is actually not strictly a good quantum number). So for a neutron in lead, the difference between the lowest shell and the Fermi level is about 5 times this, or about 35 MeV. This is a lot less than 177 MeV. Of course the lambda has a different mass and different interactions than a neutron or proton, but I don't think that's enough to make the qualitative outcome different.
If we redo the calculation for small values of $A$, we get a much bigger $\hbar\omega_0$, but $N\omega_0$ is still roughly the same.
But keep in mind that there is still some possibility that strange matter of some kind (such as strangelets) is stable. The evidence is against it, but it's not as clearcut theoretically as this example. It would be bad and scary if strange matter were more stable than normal matter. It would mean that our planet and our bodies were in a metastable state.