Para and ortho hydrogen angular momentum values In Wikipedia, it is said that:

Orthohydrogen, with symmetric nuclear spin functions, can only have rotational wavefunctions that are antisymmetric with respect to permutation of the two protons. Conversely, parahydrogen with an antisymmetric nuclear spin function, can only have rotational wavefunctions that are symmetric with respect to permutation of the two protons.

Then it seems to suggest that the symmetric wavefunctions can only have even total orbital angular momentum; while the antisymmetric ones can only have odd.
However, this seems to me to be an argument entirely analogous to the one for two-electron wavefunctions in Helium for example. But there the total orbital angular momentum can be even or odd for both symmetric and antisymmetric wavefunctions.
Why is this?
 A: You are right that the argumentation for the existence of ortho and para hydrogen is identical to that for the existence of ortho and para helium, that is, applying the permutation operator on a wave function for two identical particles should result in an eigenvalue of $+1$ or $-1$ if one considers bosons or fermions, respectively. 
Whereas in helium you apply the permutation to the electrons, in H$_2$ you interchange the hydrogen nuclei. In addition to the electronic and spin degrees of freedom, the atoms also have rotational and vibrational degrees of freedom and corresponding wave functions. After interchange of the two H nuclei, the total wave function $\phi_\text{tot}=\phi_\text{el}\phi_\text{vib}\phi_\text{rot}\phi_\text{ns}$ should by antisymmetric as hydrogen has nuclear spin $I=\tfrac{1}{2}$. 
The vibrational wave function for a diatomic molecule is always symmetric, while the symmetry of the rotational state is $(-1)^J$. The electronic ground state of hydrogen is $X\,^1\Sigma_g^+$. The symmetry of the electronic ground state is given by the product of $g$ and $+$ and therefore symmetric. As in helium, we can construct 3 symmetric and 1 antisymmetric spin functions. 
See the table below.
$
\phi_\text{tot}\qquad \phi_\text{el} \qquad \phi_\text{vib}  \qquad \phi_\text{rot}\qquad \phi_\text{ns}\\
a\qquad\quad  s \qquad \quad s \quad \quad(-1)^J \quad s\text{ or }a
$
So if you pick the antisymmetric nuclear spin function (para), the only way to make the overall wave function antisymmetric is to choose $J$ even, while for the even spin functions (ortho), $J$ should be odd. 
Note that in homonuclear molecules with zero nuclear spin, such as O$_2$, there exists only a symmetric nuclear spin function and half of the rotational states are missing. 
A: To address this part of your question:

However, this seems to me to be an argument entirely analogous to the one for two-electron wavefunctions in Helium for example. But there the total orbital angular momentum can be even or odd for both symmetric and antisymmetric wavefunctions.

The difference is that the helium atom is a three-body system, and only the two electrons are indistinguishable.  Symmetry restricts the angular momentum that the indistinguishable electrons can have relative to each other, but the elephant in the room is the nucleus.  The usual first approximation treats the nucleus as infinitely massive and fixed, and the electrons as interacting with the nucleus but not with each other.
In a diatomic molecule, on the other hand, your entire system is a pair of indistinguishable particles.  You have, effectively, a system with a single degree of freedom, and it's constrained by symmetry.
You do find electrons constrained by symmetry in positronium, an electron-positron molecule. Para-positronium can decay to an even number of photons, and usually decays rapidly to a back-to-back photon pair.  Ortho-positronium must decay to an odd number of photons to conserve parity, with the minimum number being three; the ortho-positronium decay is about a thousand times slower than para.
