Given $n$ integers $a_{1}=1, a_{2}, a_{3}, \ldots, a_{n}$, and $a_{i} \leq a_{i+1} \leq 2 a_{i}(i=1,2, \ldots, n-1)$, 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?
Let's 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 (if the sums are equal, the number can be assigned to either group). Let $\Delta_{k} \geq 0$ be the difference between the sums of the numbers in the groups after assigning $a_{k}$. We will show that $\Delta_{k} \leq a_{k}$.

Indeed, $\Delta_{n-1}=a_{n}-a_{n-1} \leq 2 a_{n-1}-a_{n-1}=a_{n-1}$. It is also clear that if $\Delta_{k} \leq a_{k}$, then $\Delta_{k-1}=\left|\Delta_{k-1}-a_{k-1}\right|$ and $-a_{k-1} \leq \Delta_{k}-a_{k-1} \leq 2 a_{k-1}-a_{k-1}=a_{k-1}$.

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