That's a very deep question. The only resolution I know is to use the Feynman path-integral in quantum theory, which evaluates the socalled propagator, which is the probability amplitude for a particle starting at ##x'## at time ##t'## and ending at ##x## at time ##t##. Up to a normalization constant it reads
$$U(t,x;t',x')=\int \mathrm{D}p \int_{(t',x')}^{(t,x)} \mathrm{D} x \exp \left (\frac{\mathrm{i}}{2 \pi} S[x,p] \right),$$
where the action functional is given by
$$S[x,p]=\int_{t'}^{t} \mathrm{d} t' [\dot{x} p-H(x,p)].$$
The integral is over all trajectories in phase space where the momentum is totally unconstrained and in position space you always have ##x(t)=x## and ##x(t')=x'##.
Now you can do a formal expansion in powers of ##\hbar##. If the action is rapidly changing the path integral will tend to vanish, because the integrand is a very rapidly oscillating sine/cosine like expression (the exponential with an imaginary argument). Thus the main contribution to the integral shoud be in the region, where the action is stationary under variations of the phase-space trajectory, and this is precisely the trajectory of the classical particle.
For a macroscopic object the approximation to take the leading order of the path integral is very good, because ##\hbar## is very small compared to the typical values of the action of the macroscopic object. That explains why we observe the particle as moving along the classical trajectory, and this explains why it is described as the stationary point of the classical action functional.