What does a symplectic vector field means in terms of the physics of a system? The mathematics of symplectic (as well as Hamiltonian) vector fields is something that has been quite clear to me for some time, but recently I have been thinking much more about what certain mathematical ideas are meant to capture from physics (as I do not have a physics background), and there is one notion that I just can't see: if some physical system admits a symplectic vector field, then what does this mean in terms of the physics of a system?  What is the physical relevance of a vector field whose flow preserves the symplectic form?
EDIT: I've come to decide that there is not really any physical meaning for a symplectic vector field, other than, perhaps, a vector field that locally models the dynamics of the system, but not globally (as $X$ symplectic implies that $\iota_X\omega$ is closed, and so the Poincaré lemma says it is locally exact; hence $X$ is locally hamiltonian, i.e. its integral curves satisfy the equations of motion).
 A: I will try to give a short (and surely partial) motivation.
A bit of notation.
Let $(M,\omega)$ be a symplectic manifold, with symplectic form $\omega$ and let $H:M\rightarrow \mathbb R$ be a smooth function (the hamiltonian). The hamiltonian vector field $X_H$ associated to $H$ is defined through
$$i_{X_H}\omega=dH.$$
An example: $(M,\omega)=(S^2, d\theta\wedge dh)$, with hamiltonian $H(\theta,h):=h$ (the height function). Then $X_H=\frac{\partial}{\partial\theta}$.
A symplectic vector field is a vector field $X$ on $M$ which preserves the symplectic form $\omega$, i.e. $\mathcal L_X\omega=0\Leftrightarrow i_X\omega$ is closed, by Cartan's magic formula.
In summary, $X$ is hamiltonian iff $i_X\omega$ is exact, while $X$ is symplectic iff $i_X\omega$ is closed. 
In general, not all symplectic fields are hamiltonian: the obstruction is given by non trivial $H^1(M)$. 
So I would see the presence of non trivial diff. geometry of the phase space as a motivation for dealing with symplectic vector fields.
For example, given the 2-torus  $(M, \omega) = (T^2, d\theta_1\wedge d \theta_2)$ the vector field $X = \frac{\partial}{\partial\theta_1}$ is symplectic but not hamiltonian.
