All s orbitals have nonzero density at the nucleus, so they overlap at least there.

There is more. The distribution ψ of different **orbitals are othogonal**, and for s orbitals, ψ is a real (non-complex) function of distance only. To obtain a zero integral over space of the product of two ψ distributions, these must change their signs at different radius.

For instance, 1s and 2s being positive near the nucleus, the zero integral of product requires 2s to become negative before 1s has decreased too much. So **significant overlapping is mandatory**. Same for 1s 3s, for 2s 3s...