= Solution
Let $F_1,\ldots,F_n$ be commuting independent integrals for $(M,H)$ and let $G_1,\ldots,G_m$ be such integrals for $(\widetilde M,\widetilde H)$. Regard each as a function on the product <phase space>. Then
$$
\{F_i,H+\widetilde H\}=\{F_i,H\}=0,
\qquad
\{G_a,H+\widetilde H\}=\{G_a,\widetilde H\}=0.
$$
The $F_i$ commute among themselves, the $G_a$ commute among themselves, and every cross bracket $\{F_i,G_a\}$ vanishes because the functions depend on disjoint canonical coordinates. Their differentials form two independent coordinate blocks, so all $n+m$ are independent on the product of the two regular sets. Thus the combined system is integrable, an instance of the <direct product of integrable Hamiltonian systems>.
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