An overkill. Let $0\leq d_1\leq d_2 \leq \cdots \leq d_n <n$ be the ordered sequence of the degrees of the graph determined by the Super-Air company (we replaced $100$ with a generic $n\geq 3$), corresponding to a sequence $v_1,v_2,\ldots,v_n$ of the vertices for which $d_i = \deg v_i$. Assume the graph is not Hamiltonian. By Chvátal's theorem, there exists $1\leq j < n/2$ such that $a_j \leq j$ and $a_{n-j} < n-j$. Since $d_i + d_{k} \leq d_j + d_{n-j} < j + (n-j) = n$ for all $1\leq i \leq j$ and $1\leq k \leq n-j$, it means for each of the $j$ vertices $\{v_1,v_2,\ldots,v_j\}$ the Concur-Air company establishes flights at most with the $j$ vertices from $\{v_{n-j+1},\ldots,v_{n-1},v_n\}$. But then the graph determined by the Concur-Air company cannot be Hamiltonian.