Solution:

If $n$ is a power of $2$ then there may be only one pile remaining; otherwise, there will be at least two piles remaining, but this can be attained. It is clear why you can reach one pile if $n$ is a power of $2$: the first $n/2$ piles can each receive pebbles from two piles with a pebble, the next $n/4$ piles can inherit two pebbles from two piles with two stones each and so forth, doubling each "generation".

Suppose $n$ is not a power of $2$ and write $n = c 2^{k}$, $c > 1$. We claim there is always a pile with a number of pebbles not divisible by $c$. This is clearly true initially. Suppose it is not true at some point, and consider what happens when you next create a pile. If this pile does not receive some pebbles from the existing pile(s) possessing a quantity not divisible by $c$, then they will maintain the invariant. If instead the pile receives some $x$ pebbles from them, then the pile will have $2x$, but $c$ does not divide $2x$ as $c$ is odd and does not divide $x$. In particular, one can never reduce to a single pile, as $n$ is of course divisible by $c$.

Finally, we show that you can always reach just two piles. This can be done via induction: Write $n = 2^{k} + r$, $r < 2^{k}$. If we get to a pile with $r$ pebbles, then the remaining $2^{k}$ pebbles can always be consolidated into one pile, irrespective of the initial distribution: we begin by having $2^{k-1}$ piles receiving $1$ pebble from two piles, but not from each other, creating $2^{k-1}$ piles with $2$ pebbles each, and then double each generation like above. To create a pile with $r$ pebbles, we have two cases. If $r < 2^{k-1}$ then simply create two piles with $2^{k-1}$ pebbles by doubling, and then take away some pebbles from both of them so that they both have $r$ pebbles, and set one of them aside. If $r > 2^{k-1}$ then create a pile with $2^{k}$ pebbles and a pile with $2^{k-1}$, and again take away some pebbles from both of them so the larger pile has $r$ pebbles. One can formulate an alternative induction argument where you use the fact $n$ can be consolidated into two piles of size $n-1$ and $1$, to show that $n+1$ can be consolidated into $n+2$ and $2$ and then show via a number-theoretic argument that you can reach $n+1$ and $1$ from here if $n$ is odd.