Let the $16$ chosen even numbers be $a_1,a_2,\ldots,a_{16}$. If we try to avoid having a sum equal to $65$ we must choose the $16$ odd numbers to be $65-b_1,65-b_2,\ldots,65-b_{16}$, where $b_1,b_2,\ldots,b_{16}$ are the other $16$ even numbers. But it must be that $\sum_{k=1}^{16}a_k= \sum_{k=1}^{16} (65 - b_k)$, i.e. $16\cdot 66 = \sum_{k=1}^{16}(a_k+b_k)= \sum_{k=1}^{16} 65 = 16\cdot 65$, absurd.