$\newcommand{\bZ}{\mathbb{Z}}$
$\newcommand{\eN}{\mathscr{N}}$
Here is a proof which does not treat overcrowding as a special case. As noted in some comments, it is really enough to prove the desired result for the case of one stone. The actually proofs takes up less space than the statements and all the notation. I will
- restate the desired result as a theorem in an equivalent but slightly more detailed way to aid the proof.
- claim a result, which implies the theorem for the case of one stone
- prove the theorem
- Show that the proof of the claim provides a much stronger result.
I will find it convenient to think of $s$ as a multiset (although I'll use the stone language too). I'll think of a move as removing a stone and replacing it with two offspring. Most of the time I will be trying to prove that $s$ is actually a set.
THEOREM: For any initial distribution of stones $s:\eN\to\bZ_{\geq 0}$ with $2 \le p =\max{s}$ and any positive integer $N$ there exists an integer $M=M(s,N)$ and a sequence of moves after which we obtain a new distribution of stones which (i) is a set (no two stones occupy the same location), (ii) has no node $n<N+p$ occupied and (iii) has no node $n>M$ occupied.
CLAIM: Consider an initial configuration $s(0)=\{{q+1\}}$ consisting of a single stone at position $q+1 \ge 2$. Let $$s(1)=\{{q+2,(q+1)^2\}}$$
$$s(2)=\{{q+3,(q+2)^2,(q+1)^2+1,(q+1)^4\}}$$ and in general $$s(k)=\{{x+1 \mid x \in s(k-1)\}} \cup \{{x^2 \mid x \in s(k-1)\}}.$$ So $s(k)$ is a multiset with $2^k$ elements, the smallest being $q+k+1$ and the largest $(q+1)^{2^k}.$ Then $s(k)$ is, in fact, a set.
Assume that we already know the $s(j)$ to be sets for all $j \lt k.$ We wish to show that $s(k)$ has no repeated elements. A number $v \in s(k)$ which is not a square can only arise as $(v-1)+1$ for $v-1 \in s(k-1).$ A value $v \lt (q+1)^2$, square or not, can only be $v=q+k+1.$ It remains to consider the square values $u^2 \in s(k)$ with $u \ge (q+1)^2.$ If $u^2-1 \notin s(k-1)$ then we know $u \in s(k-1)$ and $u \ge q+k=\min(s(k-1)).$ We will now see that, independent of $q$, if $u^2-1 \in s(k-1)$ then $u \le\frac{k+1}{2}$ so $u \notin s(k-1).$ If $u^2-1 \in s(k-1)$ then also $u^2-2\in s(k-2)$ and so on until the previous square $u^2-2u-1 \in s(k-2u+1).$ But this means $0 \le k-2u+1$ so $ u \le \frac{k+1}{2}$
This establishes the theorem for a singleton set $s=\{{p\}}$ with $M(s,N) \le p^{2^{N+1}}.$
PROOF OF THEOREM: The proof is by induction on the number of stones present. The case of one stone is the claim above with $M = p^{2^{N+1}}$. If there is more than one stone, let $s'$ be the position obtained by lowering the number of stones on the final occupied position by $1$. Then use the claim to move a single stone from that position and keep moving the offspring until all resulting stones are moved past $M(s',N).$ Then, by hypothesis, we can redistribute the (offspring of) the stones of $s'$ without colliding with any of the offspring of the initial stone.
As a consequence, if $s$ has at least two stones and $p$ is the largest filled position of $s$, then $M(s,n) \le M(\{{p\}},M(s',N)) \le p^{2^{M(s',N)+1}}.$ The strong claim below would allow smaller, but still enormous, upper bounds.
Note that the configuration $\{{p,p^2-1\}}$ leads to an immediate collision if the move is applied to both members. I will show that for a larger, but slightly thinner, configuration no collisions occur even with repeated iterations.
STRONG CLAIM: Consider, for $q+1 \ge 2,$ an initial configuration $s(0)=\{{q+1,q+2,\cdots , (q+1)^2-2\}}.$ For $k \gt 0$ let $$s(k)=\{{x+1 \mid x \in s(k-1)\}} \cup \{{x^2 \mid x \in s(k-1)\}}.$$ So $s(k)$ is a multiset with $2^k(q^2+q-1)$ elements, the smallest being $q+k+1$ and the largest $\left((q+1)^2-2\right)^{2^k}.$ Then $s(k)$ is, in fact, a set.
It is clear that $s(0)$ and $s(1)$ are sets, Assume that $1 \lt k$ and we already know the $s(j)$ to be sets for all $j \lt k.$ We wish to show that $s(k)$ has no repeated elements. A number $v \in s(k)$ which is not a square can only arise as $(v-1)+1$ for $v-1 \in s(k-1).$ A value $v \le (q+1)^2$, square or not, can only have arisen from $v-k \in s(0)$ without any squaring. It remains to consider the square values $u^2 \in s(k)$ with $u \gt (q+1)^2.$ If $u^2-1 \notin s(k-1)$ then we know $u \in s(k-1)$ and $u \ge q+k = \min(s(k-1)).$ We will now see that, independent of $q$, if $u^2-1 \in s(k-1)$ then $u \le\frac{k+1}{2}$ so $u \notin s(k-1).$ If $u^2-1 \in s(k-1)$ then also $u^2-2\in s(k-2)$ and so on until the previous square $u^2-2u-1 \in s(k-2u+1).$ But this means $0 \le k-2u+1$ so $ u \le \frac{k+1}{2}.$
Best Answer
Here is a paper about this problem: "Non-attacking queens on a triangle".
And here's another one "Putting Dots in Triangles"