Non-hamiltonian systems which evolve into hamiltonian by change of coordinates I am very new to the subject, so please forgive my very naïf question. I learned that there are some non-hamiltonian systems which can become hamiltonian, just by a change of coordinates. I was given SIR (Susceptible - Infected - Removed) model as an example: 
\begin{cases} 
\frac{dS}{dt} = - \alpha SI \\ 
\frac{dI}{dt} = \alpha SI - \beta I \\ 
\frac{dR}{dt} = \beta I
\end{cases}
with $\alpha$ and $\beta$ being real parameters.
This is clearly non-hamiltonian because it is associated with a vector field which has non zero divergence. However, by choosing $x= log(S)$ and $y= log(I)$, where $S$ and $I$ are the susceptible and infected respectively, the system becomes hamiltonian. 
What I found really strange about this result is that I'm used at seeing quantities being preserved by a change of coordinates, while here the property of the system being hamiltonian clearly isn't intrisinc: is there something deeply conceptual that I'm missing about hamiltonian mechanics?
 A: *

*Note that the divergence ${\rm div}_{\rho}X=\rho^{-1}\partial_i(\rho X^i)$ of a vector field $X=X^i\partial_i$ in general depends on a density $\rho$, cf. above comment by user mlk. The possibility of a non-trivial $\rho$ makes it more difficult to identify which 1st-order systems are potentially Hamiltonian and which are not.


*A 3D phase space can never have a non-degenerate symplectic structure, but if we ignore the last coordinate $R(t)=\beta\int^t\!dt^{\prime} I(t^{\prime})$, then we have a 2D phase space,
which always has a (local) Hamiltonian formulation, cf. this Phys.SE post.


*Concretely, the SIR-model
$$\begin{align} \dot{S}~=~& - \alpha SI~=~\{S,H\}, \cr 
\dot{I}~=~& \alpha SI - \beta I~=~\{I,H\},\cr 
\dot{R}~=~&  \beta I~=~\{R,H\},\end{align} \tag{1}$$
has non-canonical, degenerate Poisson structure
$$\begin{align}    \{S,I\}~=~&SI, \cr \{I,R\}~=~&\frac{\beta}{\alpha}I, \cr \{S,R\}~=~&0,\end{align} \tag{2} $$
and Hamiltonian
$$ H~=~\beta \ln S  -\alpha (S+I) .\tag{3} $$


*If we follow OP's suggestion to change coordinates
$$\begin{align} s~:=~&\ln S, \cr
  i~:=~&\ln I , \cr
 r~:=~&R, \end{align}\tag{1'}$$
then the fundamental Poisson brackets (2) become constant
$$ \begin{align}   \{s,i\}~=~&1, \cr \{i,r\}~=~&\frac{\beta}{\alpha}, \cr \{s,r\}~=~&0, \end{align}\tag{2'} $$
and it becomes obvious that the Jacobi identity is satisfied as it should. The Hamiltonian reads
$$ H~=~s -\alpha (e^s+e^i) \tag{3'} $$
in the new coordinates.
