To solve this problem, we need to determine the smallest number of nonempty piles of pebbles that can be obtained when starting with $n$ piles, each with one pebble, by performing a series of moves. In each move, we choose two piles, take an equal number of pebbles from each, and form a new pile with these pebbles.

Step 1: Understanding the Moves

Each move involves:
- Selecting two piles, say with $a$ and $b$ pebbles each.
- Removing $k$ pebbles from each pile ($1 \leq k \leq \min(a, b)$).
- Forming a new pile with the $2k$ pebbles removed.

The total number of pebbles in the system remains constant at $n$. After such a move, the number of piles changes as follows: two piles are reduced by $k$ pebbles each, and a new pile of size $2k$ is created.

Step 2: Characterizing Optimal Configuration

When $n$ is a power of 2, it is possible to perform moves such that all pebbles are eventually concentrated into a single pile. This can be achieved through successive pairings:

- Start with $2$ piles of $1$ pebble and combine them to get $1$ pile with $2$ pebbles.
- Now pair this $2$ pebble pile with another $2$ pebble pile to get a pile of $4$ pebbles.
- Continue doubling in this way until all pebbles are in one pile.

Thus, for such cases:
$$\boxed{1}$$

When $n$ is not a power of 2, pairing cannot result in a single pile because repeatedly dividing by 2 will not terminate at 1. In this situation, regardless of the moves, we are unable to end up with just a single pile, but we can reach a state with exactly two piles:

- Use binary representation: every positive integer can be expressed as a sum of distinct powers of 2.
- If $n$ is not a power of 2, this decomposition into powers of 2 has more than one term.
- Therefore, we can make moves such that pebbles eventually consolidate into two piles, each representing the largest possible sum of grouping the binary terms.

Hence, when $n$ is not a power of two, the minimum number of piles is:
$$\boxed{2}$$

Final Solution

The smallest number of nonempty piles possible, depending on whether $n$ is a power of 2 or not, is given by:
$$\boxed{ \begin{cases} 1 & \text{if }n\text{ is a power of }2 \\ 2 & \text{otherwise} \end{cases} }.$$