16.15. Given $n$ integers $a_{1}=1, a_{2}, a_{3}, \ldots, a_{n}$, and $a_{i} \leqslant a_{i+1} \leqslant 2 a_{i}(i=1,2, \ldots, n-1)$, and the sum of all numbers is even. Can these numbers be divided into two groups such that the sums of the numbers in these groups are equal?
16.15. Answer: Yes, it is possible. We will assign the number $a_{n}$ to one group and the number $a_{n-1}$ to the other. Then we will sequentially assign the numbers $a_{n-2}, a_{n-3}, \ldots, a_{1}$ to the group with the smaller sum of numbers (if the sums are equal, the number can be assigned to either group). Let $\Delta_{k} \geqslant 0$ be the difference between the sums of the numbers in the groups obtained after assigning $a_{k}$. We will show that $\Delta_{k} \leqslant a_{k}$. Indeed, $\Delta_{n-1}=a_{n}-a_{n-1} \leqslant 2 a_{n-1}-a_{n-1}=a_{n-1}$. It is also clear that if $\Delta_{k} \leqslant a_{k}$, then $\Delta_{k-1}=\left|\Delta_{k}-a_{k-1}\right|$ and $-a_{k-1} \leqslant \Delta_{k}-a_{k-1} \leqslant a_{k}-a_{k-1} \leqslant$ $\leqslant 2 a_{k-1}-a_{k-1}=a_{k-1}$.

After we have distributed all the numbers into two groups, we will obtain groups with sums $S_{1}$ and $S_{2}$, and $\left|S_{1}-S_{2}\right| \leqslant a_{1}=1$. By the condition, the number $S_{1}+S_{2}$ is even, so $S_{1}=S_{2}$.