Given a finite number of stones. We place every stone on an integer. Prove that, given different movements, we can only make finite number of moves

Suppose that there is a counterexample, i.e. an infinite set of moves from some configuration with a finite number of stones, $S$ say. If $S=1$, then no moves are possible and so we can suppose that we have a counterexample with $S$ minimal.

If the sequence contained a move of Type 1 then, after that move, there would only be $S-1$ stones and, by minimality of $S$, there could only be a finite number of further moves. All moves are therefore of Type 2 and, as in the post, this means that the total weight $W$ is unchanged for $x=\frac{1+\sqrt{5}}{2}$. A stone on position $n$ contributes $x^n$ to $W$ and, in consequence, the integer values for any occupied position is bounded above by the log of $W$ to the base $x$.

Since we are dealing with integers, this upper bound, $m$ say, is attained. Now consider moves from a configuration with at least one stone at position $m$. No move of Type 2 from this configuration can involve stones on $m$ and so the remaining moves involve at most $S-1$ stones. By minimality of $S$ there can only be a finite number of further moves after all.