I think the following is probably a trivial application of KAM theory, which I know little about, so I am hoping someone can give me an answer pointing me in the right direction.
Suppose I have a Hamiltonian system which is just a collection of $N$ non-interacting harmonic oscillators. The Hamiltonian is $H_0=\sum^N_{i=1}\omega_i J_i$, where $J_i$ are action variables with some conjugate angle variable $\theta_i$ that doesn't appear in the Hamiltonian so all the $J_i$ are trivially conserved. Further suppose the frequencies $\omega_i$ are non-resonant, i.e. for any non-zero set of integers $n_i$ we have $\sum_{i} n_i \omega_i\neq 0$.
Is the perturbed Hamiltonian $H(J,\theta;\lambda)=H_0(J)+\lambda V(J,\theta)$ also integrable for small enough $\lambda$?
I originally asked this question on Physics Stack Exchange, and gave a sketch of an argument that we could construct conserved quantities $J'_i$ to any order in $\lambda$ that are perturbations of the $J_i$ (see here). But I don't know if I can trust this reasoning and I am really looking for some kind of non-perturbative understanding. My hand-wavy understanding of KAM theory is that the resonant tori are 'destroyed' for arbitrarily small $\lambda$ and then non-resonant tori that are 'close' to the resonant ones are progressively destroyed as $\lambda$ is increased. So I thought perhaps a system with no resonances at $\lambda=0$ maintains integrability for some finite interval of $\lambda$.