Why doesn't the deuterium nucleus have spin $0$? A deuterium nucleus is composed of a proton and a neutron. Both have spin $\tfrac12$ so I would expect the deuterium to have two possible spins: $1$ for the triplet and $0$ for the singlet. But apparently deuterium always has spin $1$ and the spin $0$ state doesn't exist. Why?
 A: Deuterium is light enough that isospin is a good symmetry: it's fair to neglect the electric charge and consider only the strong interaction between the proton and neutron.  In that case we should expect essentially the same excitation structure in the diproton, the dineutron, and the neutron-proton system.
The proton and neutron are both fermions, and a state containing two of them must be antisymmetric under exchange.  If they are bound without any orbital angular momentum, the only way to make the state antisymmetric is for the two particles to occupy a spin singlet. So the ground state of the diproton or dineutron must have spin zero. Since the diproton and dineutron are both unstable, isospin symmetry tells us that spin-zero deuteron should also be unstable.
In the stable deuteron the state is made antisymmetric by the isospin part of the wavefunction: the deuteron is a (symmetric) spin triplet but an (antisymmetric) isospin singlet.  (It's true but irrelevant that the deuteron wavefunction is only mostly $s$-wave; there's a small contribution of $d$-wave with two units of orbital angular momentum, but that doesn't change the symmetry arguments or the total nucleon spin.)
If you prefer, you can turn this argument around. If it's the case that the strong interaction is more important than electrical repulsion in light nuclei, and if you found a stable two-nucleon state with zero angular momentum, you would expect to find that two-nucleon state for all allowed values of the charge: the dineutron, the deuteron, and the diproton.  We don't find any evidence for stable dineutrons or diprotons, and we also don't find any spinless bound deuterons.
A: I am bit confused by Rob's reasoning above, he does not talk about the parity of the strong wave function and the color degree of freedom and confinement which seems to me the key of the argument. Please let me know if there is any mistake about my reasoning below.
This is how I would argue. The wavefunction of the deuteron is the tensor product of four wavefunctions, each representing a particular degree of freedom:
$$\psi=\psi_{spatial}\psi_{spin}\psi_{flavour}\psi_{color}$$
We can assume that the strong interaction (QCD) is the only one relevant here and neglect all others, from the point of view of QCD a proton and a neutron are basically identical (and having have-integer spins are of course fermions) therefore we must require the total wave function $\psi$ to be antisymmetric under the exchange. Now:
Assume $l=0 \Rightarrow \psi_{spatial}$ has parity $+1$.
Since QCD does not distuinguish the flavour degree of freedom of the particles the flavour wave function is symmetric $\Rightarrow \psi_{flavour}$ has parity $+1$.
This is the key point, it looks like QCD only allows bound states that are singlets of color (confinement) singlets of color have parity $-1$, color is the name that was given to the strong charge and comes in three varieties: red, green, blue. So $\psi_{color}$ has parity $-1$.
As explained above we need $\psi$ to be antisymmetric, so we must require $\psi_{spin}$ to be symmetric. We know that two particles of spin $\frac{1}{2}$ can only couple to states of total spin $0$ or $1$ but of those two only the latter is symmetric. That is why the deuteron has spin 1. 
A: The reason why deuterium occurs only with spin 1 is that nucleon-nucleon interactions are spin-dependent, and the energy scales are such that the state with anti-parallel spins ($S_z=0$) is energetically unstable. This little energy balance difference actually has very far reaching consequences for the structure of nuclei... and, as a consequence, for the structure of our world as a whole.
The explanation in terms of isospin provdied by @rob is a common way to describe this asymmetry of nuclear forces. However, it si worth keeping in mind that isospin is not a real quantul number (as @rob also notes in their first sentence, saying that it is a godo symmetry for this particular case).
A: Proton has a positive magnetic moment, whereas neutron has a negative one. As a matter of fact, there will be a small attractive contribution if their spins are parallel (and their magnetic moments antiparallel). Consequently, the lowest energy state of a proton-neutron system is the parallel spin state. The parity of deuteron is positive, thus orbital momentum should be either 0 or 2 and total spin is either $$l+s_p+s_n=0+1/2+1/2=1,$$ or $$l-s_p-s_n=2-1/2-1/2=1$$
Actually it is a mixture of both states, 96% with $l=0$ and 4% with $l=2$, as it comes out from the comparison of total magnetic moment of deuteron with the sum of proton and neutron magnetic moments. They differ considerably, that being the consequence of the contribution of the $l=2$ state. 
