The answer is all $n$ of the form $4k+2$. For convenience of discussion, regard the stars from left to right as occupying position 1 through position $n$. We discuss the following cases separately.

1. Even $n$:
Note that in the case of even $n$, Becky can always choose to take all the stars in odd positions (by first taking the star at position 1, and afterwards always taking from the same side as Anya), or take all the stars in even positions (by first taking the star at position $n$, and afterwards always taking from the same side as Anya). Let $O$ be the sum of the values of the stars in odd positions, and $E$ be the sum of the values of the stars in even positions, then $O+E = n(n+1)/2$.

- When $n = 4k + 2$, $n(n-1)/2$ is odd, so Becky only needs to choose whichever of $O$ and $E$ has the larger sum and take that one.
- When $n = 4k$, note that Becky can never lose (because in the worst case she can obtain the same amount as Anya). And if Anya initially arranges the stars as
$$1, 3, 5, \dots, 4k-1, 4k, 4k-2, \dots, 2$$
and always takes from the same side as Becky, then in the first $k$ rounds Anya can guarantee taking a value 2 more than Becky each time, and in the last $k$ rounds Anya, in the worst case, takes a value only 2 less than Becky each time, so Anya also can never lose, hence the two players draw.

2. Odd $n$:
If $n = 2k-1$, Anya only needs to initially place the stars with values 1 through $k$ in the odd positions, and place the remaining stars in the even positions, and always take from the same side as Becky, then Anya can guarantee taking all the stars from $k+1$ through $2k-1$, whose total value is $3k(k-1)/2$, which is greater than the value $k(k+1)/2$ that Becky obtains, so Anya has a winning strategy.