Is it possible for the Action $S$ to *not* have a stationary point? So the path of an object in configuration space is given by Hamilton's principle, which states that the path which the particle travels on is the one on which the action is stationary:
$$\delta S = \delta \int L dt = 0.$$
I was thinking though, is it possible for the stationary point of the Action to not exist? If it doesn't, does this mean the particle doesn't move?
 A: We know that if the Euler-Lagrange equation holds, then the action must have a stationary value. The Euler-Lagrange equation is
$$\frac{\text d}{\text d t}\frac{\partial L}{\partial\dot q}=\frac{\partial L}{\partial q}$$
For generalized coordinate $q$ and Lagrangian $L$.
If we take $L$ to be $L=T-V$ for kinetic energy $K=\frac12mv^2=\frac12m\dot x^2$ and potential energy $V=V(x)$ then we have
$$\frac{\text d}{\text d t}\frac{\partial L}{\partial\dot x}=m\ddot x=ma$$
and
$$\frac{\partial L}{\partial x}=-\frac{\partial V}{\partial x}=F$$
And so we end up with Newton's second law $F=ma$
So if we want Newton's laws to hold$^*$, we must also have that the Euler-Lagrange equation holds for our Lagrangian $L=T-V$, which means there must be a stationary point for our action.
In other words, if we don't have a stationary point, then we aren't dealing with physically realizable trajectories.

$^*$Of course, the power of using Lagrangian or Hamiltonian mechanics is that we have more freedom in using generalized coordinates instead of just considering spatial coordinates, but it still doesn't mean we throw out Newton's laws.
A: *

*A generic functional does not have to have stationary points. 
For an elementary (although admittedly somewhat artificial) example take e.g. a Lagrangian$^{\dagger}$ $$L=F_{\rm ext}y, \qquad F_{\rm ext}~\equiv~{\rm constant}~\neq~ 0, $$ for a uniform external force field $F_{\rm ext}$. It does not have a stationary point. The corresponding Lagrange equation $F_{\rm ext}=0$ is never satisfied and has no stationary solutions. This is true even if we impose, say,  Dirichlet boundary conditions $y(t_i)=0=y(t_f)$. 

*Of course, if we assume that the action $S$ implements Newton's 2nd law in a well-posed physics problem, then it should have a stationary point.
--
$^{\dagger}$ Imagine e.g. that there is no time, i.e. we are studying the static problem, or that we consider the ideal limit of vanishing mass $m= 0$ and hence vanishing kinetic energy $\frac{1}{2}m\dot{y}^2= 0$. 
A: A model is defined by a system of equations, the equations of motion, that specify how the various players are allowed to behave. The whole system of equations of motion can often be encoded in the single equation $\delta S=0$, where $S$ is the integral of a suitably chosen lagrangian $L$. This is usually what we mean by "the action" for a given system.
If the action doesn't have stationary points, then it isn't useful for describing the behavior of any system in that way. Conversely, if we have a system whose equations of motion can be encoded in the form $\delta S=0$, then — by definition — the action has all the stationary points we need for describing that system's behavior. Of all the behaviors we can imagine, only those that satisfy $\delta S=0$ are actually allowed.
Hamilton's principle implicitly assumes that we are dealing with a system whose equations of motion can be encoded in the form $\delta S=0$ and that the lagrangian $L$ was chosen to implement this encoding. Only behaviors that satisfy $\delta S=0$ are allowed. Other behaviors will not occur. If "the particle doesn't move" is an allowed behavior, then this behavior will satisfy the condition $\delta S=0$; it will be one of the stationary points of the action.
