Is there a proof that shows that the total Hamiltonian for a non-interacting system is the sum of Hamiltonians of the components? According to "Modern Quantum Chemistry" by Szabo and Ostlund, the Hamiltonian for a system of non-interacting electrons can be written as $$\mathcal{H}=\sum^N_i\mathcal{h(i)}$$
here $\mathcal{h(i)}$ is the one electron Hamiltonian. $\mathcal{h(i)}$ consists of a kinetic energy term for electron $i$ and the potential felt by electron $i$. 
Why can we add one electron Hamiltonians to get the Hamiltonian for the complete system?
In addition, the page about Hamiltonians in wikipedia 

the result is that the Hamiltonian of the system is the sum of the separate Hamiltonians for each particle

Intuitively, this makes sense. But is there an mathematical proof? Is there an "if and only if" proof? 
 A: There can't be a "proof", since this just the hamiltonian choosen to give the right results: In QM, you START with the hamiltonian, which then is supposed to describe your system. It's always "Whoo, this is the Hamiltonian, let's see how my system behaves. Oh, it behaves the same way I can measure stuff. This must be the right Hamiltonian". 
However, there are ways to make it plausible to use a certain Hamiltonian. For a single particle you want to reproduce classical mechanics in the limit of $\hbar$ being small. So you take the Hamilton-Function, you've been used to, and replace position and momentum with Operators. 
In classical mechanics, for serveral noninteracting particles, you just add the hamilton-functions, because then every particle, now described by the multi-particle-hamilton-function, will behave according to the same equations of motion. And it's the same with Quantum-mechanics:
You want a Hamiltonian, according to which the particles behave independently of each other (that would be the non-interacting hamiltonian by definition), and in the same manner they would act when they were the only particle. The only way to achieve that is to add all the Hamiltonians of the single particles. 
A: I'll write about classical Hamiltonian systems for now. Maybe add later about quantum theory.
Knowing Hamiltonian equation of motion for some system means that you know derivatives of the Hamiltonian,
\begin{equation}
\frac{\partial H}{\partial P_I}=F_I(P,Q),\quad \frac{\partial H}{\partial Q_I}=G_I(P,Q)
\end{equation}
You can then recover the Hamiltonian up to the constant by path-independent integration,
\begin{equation}
H=\int\limits_{(P_0,Q_0)}^{(P,Q)} \sum_I\Big(dP_I\,F_I(P,Q) + dQ_I\,G_I(P,Q)\Big)
\end{equation}
That two subsystems with canonical variables $(P_I,Q_I)=\begin{pmatrix}(p_i,q_i)\\(\tilde{p_j},\tilde{q_j})\end{pmatrix}$ don't interact means that $(\dot{p}_i,\dot{q}_i)$ don't depend on $(\tilde{p}_j,\tilde{q}_j)$ and vice versa. From Hamiltonian equation of motion that requires,
\begin{align}
\frac{\partial H}{\partial p_i}=F_i(P,Q)=f_i(p,q),\quad \frac{\partial H}{\partial q_i}=G_i(P,Q)=g_i(p,q),\\
\frac{\partial H}{\partial \tilde{p}_j}=F_j(P,Q)=\tilde{f}_j(\tilde{p},\tilde{q}),\quad \frac{\partial H}{\partial \tilde{q}_j}=G_j(P,Q)=\tilde{g}_j(\tilde{p},\tilde{q})
\end{align}
The above integral then separates into two terms,
\begin{equation}
H=\left[\int\limits_{(p_0,q_0)}^{(p,q)}\sum_i\Big(dp_i\,f_i(p,q)+dq_i\,g(p,q)\Big)\right] + \left[\int\limits_{(\tilde{p}_0,\tilde{q}_0)}^{(\tilde{p},\tilde{q})}\sum_j\Big(d\tilde{p}_j\,f_j(\tilde{p},\tilde{q})+d\tilde{q}_j\,\tilde{g}_j(\tilde{p},\tilde{q})\Big)\right]
\end{equation}
