That's a really nice question :)
The main equation that is being solved here is:
$$\rho \frac{\partial^2 \vec{u}}{\partial t^2}-\textrm{div}(\sigma(\vec{u}))=\vec{f}\tag{1}$$
with $\sigma(\vec{u})$ the elastic tensor that describes the behaviour of the material that is being considerated. This is given by a constitutive law. And $\vec{f}$ the internal forces.
This equation has (as is natural) two kind of BC:
$$\vec{u}=\vec{u}_D \quad \forall \vec{x}\in\partial \Omega\qquad \sigma\vec{n}=\vec{t}\quad \forall \vec{x}\in\partial\Omega\tag{BC}$$
that are imposed in the Dirichlet boudary where the displacements are fixed and in the Neumann boundary where tractions are known, respectively. There are as well two initial conditions, but they are not so important here...
In dynamics (equation $(1)$ you can impose whatever, because it is not a requirements that the body is at rest... but what happens if the body does?
This is statics... and its behaviour is described by the next equation:
$$-\textrm{div}(\sigma(\vec{u}))=\vec{f}\tag{2}$$
subjected to the same BC as before $(BC)$.
Now everything change...
I will ask you using:
Physics
Could a body be at rest without moving? The answer is yes!
If the resultant of the applied forces adds to 0, then the body is in equilibrium. For this to happen, there is no need to specify any Dirichlet BC but in one node (to specify the spatial reference).
Therefore, equilibrium of forces must be present:
$$\vec{F}=\int_{\Omega}\vec{f}\,dV=-\int_{\partial\Omega}\vec{t}\,dS=\vec{T}\tag{*}$$
This means that the resultant of the body forces $\vec{F}$ must exert a force equal but in oposite direction respect to the external forces that are applied to the body (as a boundary condition).
If this occurs we don't have anything to be afraid of.
For example, a rope, whose extrema are being equally pulled does not remains at equilibrium even the fact that in their boundaries two traction BC are imposed. Of course we must specify in which spatial point the rope is, otherwise there would be infinite solutions to the problem in which the rope would be translated in space. In this example, if the rope is hung under gravity, in the lower extremum the BC would be $\vec{t}_l=\sigma \vec{n}=\vec{0}$ (no applied force), while in the upper $\vec{T}=\vec{t}_u-\vec{t}_l=\vec{F}=\int_0^{L}\rho \vec{g}\,dz$ (force exerted by the whole mass that it supports). This problem will have a unique solution if one point of the rope is known, otherwise the rope could be in equilibrium wherever.
Mathematics
This is so easy:
Apply the Gauss Th. to $(2)$ and apply the $(BC)$. If to Dirichlet BC are imposed, we obtain the restriction:
$$\int_{\Omega}\left[-\textrm{div}(\sigma(\vec{u}))\right]dV=-\int_{\partial\Omega}\sigma\vec{n}\,dS=-\int_{\partial\Omega}\vec{t}\,dS=\vec{T}=\int_{\Omega}\vec{f}\,dV=\vec{F}$$
Which is the same equation deduced above.
Edit
Thanks to @origimbo to remember me the rotations!
Of course, apart from the equilibrium of forces, there must be an equilibrium of torques! Therefore, the provided boundary conditions must be consistent with this equilibrium:
$$\int_{\Omega}\vec{x}\times\vec{f}\,dV=-\int_{\partial\Omega}\vec{x}\times\vec{t}\,dS\tag{**}$$
with respect to an arbitrary origin of coordinates.
Also, this last statement may be deduced taking the cross product of $(2)$ with an arbitrary vector $\vec{x}$, integrating by parts and noting that $\sigma = \sigma^{T}$.
If the provided $BC$ (neumann only) given by $(BC)$ fulfills $(*)$ and $(**)$ then your problem is well posed once fixed one node.
