Let $m < n$ be positive integers. Start with $n$ piles, each of $m$ objects. Repeatedly carry out the following operation: choose two piles and remove $n$ objects in total from the two piles. For which $(m,n)$ is it possible to empty all the piles?
I think it is the case that for $k$, natural number, $(k,2k), (ak, ak+a)$ works.
Observation: $m\ge n/2$.
Non-trivial solvable case: $(m,n)=(9,12)$
The first solution is trivial. I proved the second one by induction.
Are there any other solutions and is the second solution correct?
The answer is the set of pairs $(m,n)$ such that $n>m$ and $n-m\mid m$. First, let $m=(n-m)k$, where k is a positive integer. It can be rewritten as $m(k+1)=nk$. Since $k$ and $k+1$ are always coprime, $k\mid m$ and $k+1 \mid n$. If we denote gcd$(m,n)$ by $a$, we get the answer in your form $(ak, ak+a)$.
Now let us prove that all such pairs work. We have $k+1 \mid n$, so we divide all the piles in groups of $k+1$. Each group can be processed as follows: we take the first pile ($m$ objects) and $1/k$-th part of the last pile in the group (remember that $k\mid m$), then the second pile and again the part of the last pile, and so on.
Finally, let us prove that $n-m\mid m$ is a necessary condition. Assume that at some step we took $k$ objects from a pile, where $n-m \not\mid k$. What did we take from the second pile? $n-k$ objects. Since there are $m$ objects in any pile, the fate of the other $m-(n-k)=k-(n-m)$ is still unknown. Note that $k-(n-m) \not\mid k$. All these objects were taken at some point, perhaps in several steps. But there was such a step when we took the number of objects from these $k-(n-m)$ which is not divisible by $n-m$. We can repeat this reasoning and every time the number of objects will get smaller but would never be divisible by $n-m$. So it can’t get to $0$ which it should.
Therefore, number of the objects we take from a pile is divisible by $n-m$, and so is the number of the objects in the whole pile. We are done.