In a row are 23 boxes such that for $1\le k \le 23$, there is a box containing exactly $k$ balls. In one move, we can double the number of balls in any box by taking balls from another box which has more. Is it always possible to end up with exactly $k$ balls in the $k$-th box for $1\le k\le 23$?
We show by induction that it's possible to get any permutation of the set $1, 2, \dots, n$ by using the procedure described in the problem.
If $n=1$ or $n=2,$ the statement is clearly true.
Now assume that the claim is correct for all $k < n.$ Then we get any permutation of $1, 2, \dots, n-1,$ and it suffices to show that we get a permutation of $1, 2, \dots, n$ where $n$ changed its position.
If $n$ is even, then $n = 2m$ for some positive integer $m.$ Then we can apply the procedure to the numbers $n$ and $m.$
If $n$ is odd, then $n = 2m + 1$ for some positive integer $m.$ By applying the procedure to $2m+1$ and $2m,$ we get $4m$ and $1.$ Next we use $4m$ and $2m-1,$ and get $4m - 2$ and $2m + 1.$ Then we use $4m - 2$ and $2m - 2,$ and get $4m - 4$ and $2m.$ Continuing in the same fashion, in the end we will get a permutation where $2m+1$ changed its position.