Answer: $n=2^k$ for all non-negative integers $k$ .

Solution: We will show that if $n$ is a power of $2$ then eventually each person will have exactly one apple, and if $n$ is not a power of $2$ this will not happen.

Assume that $n=2^k$ . We claim that after $2^k-1$ steps, everyone will have exactly one apple. We proceed by induction on $k$ . The base case $k=0$ is trivial. For the induction step, assume that the result holds for $k$ and we will show it holds for $k+1$ . Notice that for each of the first $n-1$ steps the range of people that have an apple will expand by one in the clockwise direction. Thus no apple will make its way around the circle in the first $n-1$ moves, so we can imagine "cutting" the circle to the left of Ali and only considering the passing in straight line. By the inductive hypothesis, after $2^k-1$ moves, Ali will be left with $2^k+1$ apples, the $2^k-1$ friends to the right of Ali will have exactly $1$ apple, and no one else has apples yet. After $1$ more step, Ali will be left with $2^k$ apples, the friend $2^k$ spots to the right of Ali will also have $2^k$ apples, and no one else will have apples. Thus by the inductive step, after another $2^k-1$ steps all $2^{k+1}$ people will have exactly $1$ apple. This completes this part of the solution.

Now we prove two claims.

Claim 1. All such $n\neq 1$ are even.
Proof. Assume $n\neq 1$ works. Consider the situation one step before everyone gets an apple. Everybody having at least one apple must have exactly $2$ apples in order to end up with just $1$ apple after the step. Then $2|n$ , as claimed.

Claim 2. If $n=2k$ works, then $n=k$ also works.
Proof. Consider the party with $n=2k$ people. Let $A$ denote the set of $k$ people which are an even number of seats away from Ali and let $B$ denote the set of the other $k$ people. We claim that after every two steps, only the people in $A$ will have apples, and each of them will have an even number of them. Additionally, the people in $A$ function as a party of $k$ people where every two steps it is as if they pass with $2$ apples instead of $1$ . Notice that this is true from the beginning. Now consider a person in $A$ that has no apples and is adjacent (to the left) to a block of friends in $A$ with apples. After the first step all the people in $B$ in front of a person from the block will receive $1$ apple. The person to the left of the block still does not have an apple so after the second move all the apples received by people in $B$ plus one additional apple from each person from the block of friends in $A$ will go to the person in consideration. Thus effectively, after two steps, the people in $B$ just helped "passing" the apples and returned to having no apples, while the people in $A$ functioned as a sub-party with $k$ people and twice as many apples. This only stops when everyone in $A$ has exactly $2$ apples, in which case we can not consider a person in $A$ that has no apples and thus after one more step, everyone will have an apple. But by examining our sub-party, this means that $n=k$ must also work, as claimed.

Now if $n$ is not a power of $2$ , then express $n$ as $2^k\cdot m$ for a non-negative integer $k$ and an odd integer $m\geq 3$ . By Claim 2, if $n$ works then $m$ must also work. But by Claim 1, this is a contradiction, as desired.