For hydrogen, other than in the 1s state, the electron wavefunctions have radial and/or angular nodes where the electron probability density is zero.
In helium or further atoms with more than one electron, considering correlation of electrons, what happens to these nodes? Are there still radial and angular nodes where electron probability density is zero?
For example, if helium is in a 2s2s state, is there anywhere that an electron cannot be?
My answer to this question is the following. Nodes can happen for highly excited states, whose average distances are typically much larger than the Bohr radius. Consider Na, Cs etc, which have one outermost shell electron. When the outermost shell electron is highly excited, the case is not much different from that of a hydrogen atom (i.e. the indistinguishability of electrons may be ignored when electrons are well separated.) Thus there can be nodes in these highly excited states.
A more non-trivial question is, whether the total electron density can have nodes for the ground state. I tend to think that it's not possible. My following naive, sketchy explanation may be seriously wrong. For a second-order eigenvalue problem, we have the sturm theorem (?; in doubt), which guarantees that the roots of successive states occur in an interlacing manner. That means, where state $i$ has a node at $r_i$, state $i+1$ cannot have a node at $r_i$. Thus the density, which is equal to the sum of square of wave function of all the occupied states, cannot be zero at one point. We have a similar second-order eigenvalue problem for each angular momentum $l$, and so the argument goes.
The above too sketchy, so let's formalise it a bit based on the one-particle approximation. The total ground-state electron density is proportional to $ \sum_{\{n,l, m, s\}; \textrm{occupied}}|\psi_{nlm}(r)|^2 |Y_{lm}(\theta, \phi)|^2$. Here $s$ is for the spin, $n$ labels the energy state (1 for ground, 2 for first excited, etc), and the sum is over all occupied states. Note that in order for the electronic density to be zero at $r_0$, each term has to vanish at $r_0$. But remember that for $l=0$, $m=0$, the $n=1$ radial wave function $\psi_{100}(r)$ can have no node, and $Y_{00}$ is a constant, which means the term $\psi_{100}(r)Y_{00}$ can vanish nowhere in space. This state has to be occupied in the ground state.
There is a proof of node for the triplet-S state of helium in What Do We Know About Wavefunction Nodes? Recent Advances in Computational Chemistry vol. 2. (alternative link to preprint version):
The wavefunction can be written as a function of only 3 coordinates: the distance ($r_1$) of the first electron from the nucleus, the distance ($r_2$) of the second electron from the nucleus, and the distance ($r_{12}$) between the electrons.
$\Psi = \Psi(r_1, r_2, r_{12})$
by the Pauli exclusion principle:
$\Psi(r_1, r_2, r_{12}) = -\Psi(r_2, r_1, r_{12})$
Then, suppose both electrons are the same distance ($r$) from the nucleus, but not necessarily at the same point:
$\Psi(r, r, r_{12}) = -\Psi(r, r, r_{12})$
Therefore,
$\Psi(r, r, r_{12}) = 0$
In other words, there is a node defined by $r_1 = r_2$.
The two electrons can never be the same distance from the nucleus.
The reference goes on to say that in 6-dimensional space, the node is a 5-dimensional hypersurface.