The answer is all $n \equiv 2 \pmod 4$. In a particular instance of the above game, let $n_1, n_2$ denote the sum of star values of Becky and Anya. We then define the score of that game as the difference \[n_1-n_2\]We change the objective from wanting to keep the above difference $> 0$ or $\leq 0$ (depending on whichever player) to instead maximising or minimising the above score (depending on whichever player). (note that this does not harm the winning strategy of either player, since if it is possible to win the game with the earlier strategy, it must also be possible to win the game with the new strategy) We first prove the following Lemma: Lemma: Consider a game between the two players on the ordering \[\ell, \ell+1, \dots, r\](Becky starts) for some two positive integers $\ell, r$. The score after an optimal game is then either $\frac{r-\ell+1}{2}$ or $\frac{r+\ell}{2}$, depending on whichever is an integer. Proof: We induct on the length of the ordering $r-\ell+1$. The base case of an empty ordering is clear. We now split into cases depending on the parity of $r-\ell+1$, the number of numbers in the order. Case 1: $r-\ell+1$ is even. In this case, Becky either removes $\ell$ or $r$. From the inductive hypotheses, we know that the optimal score on the games $[\ell+1, r]$ and $\ell, r-1$ are $-\frac{r+\ell+1}{2}, -\frac{r+\ell-1}{2}$ (the negative signs show up since in these games, Anya starts). Therefore Becky must choose the maximum of \[\ell - \frac{r+\ell+1}{2}\]and \[r - \frac{r+ \ell-1}{2} = \frac{r-\ell+1}{2}\]which can easily be seen to be the latter. This proves the inductive hypotheses. Case 2: $r-\ell$ is even. Again, Becky removes either $\ell$ or $r$. From the inductive hypotheses these lead to games with scores: \[\ell - \frac{r- \ell}{2}\]or \[r - \frac{r-\ell}{2} = \frac{r+\ell}{2}\]The latter is larger, hence Becky chooses that strategy. This proves the inductive hypotheses. To finish, the Lemma is proven now. Now let us return to the problem at hand. For $n = 2k+1$ odd, consider the following order \[2, 3, \dots, 2k+1, 1\]We show that Anya can make the score $\leq 0$, in particular, Becky cannot win here. Anya follows the following strategy: while there are at least two elements before $2k+1$ and Becky removes the left most element, Anya responds with removing the right most element. On the other hand, if Becky removes $1$ at any point, then Anya removes the $2k+1$ and reduces the game to some interval game, and plays that acoording to the strategy prescribed in the Lemma. We first dispose of the case when Becky does not act on the $1$ at all. Then the score decreases by exactly one every pair of elements exchanged on the left end, and in the end the score will be $-k$ with only the number $1$ left on the board. Becky takes this for herself, making the score $-k+1 \leq 0$. Now suppose that Becky acts on the $1$ after $x < k$ pairs of elements are removed from the left end. Thus the score currently is $-x$ with the numbers on the board being \[2x+2, \dots, 2k+1, 1\]After $2k+1, 1$ are removed, the score becomes $-x-2k$ with the state being \[2x+2, \dots, 2k\]We know that an optimal game on this state leads to a score of $\frac{2x+2+2k}{2} = x+k+1$. Adding this in with the previous score we obtain a final score of \[-x-2k + x + k + 1 = 1-k \leq 0\] This shows that Becky cannot always ensure a win for $n$ odd. Now let $n = 4k$ be a multiple of $4$, and consider the following ordering: \[1, 3, 4, 2, 5, 7, 8, 6 \dots, 4k-3, 4k-1, 4k, 4k-2\]We show that this ends in a draw. Form the following pairs $(1, 3) ; (4, 2) ; (5, 7) \dots (4k-3, 4k-1) ; (4k, 4k-2)$. We make Anya follow the following strategy: whenever Becky removes on element from one of these groups, Anya removes the other. We inductively show that Anya's strategy prevents Becky from winning. If a pair above is removed from the left (i.e., the left element is removed first), we mark it with a $L$, and mark it with $R$ in the other case. Note that pairs like $(4m-3, 4m-1)$ removed from the left increase the score by $2$, whereas other pairs like $(4m-2, 4m-4)$ decrease by the same. On the other hand, when removed from the right, these score changes switch signs again. Now note that every turn, either the leftmost or the rightmost is marked with an $L$ or a $R$. Since there are $4k/2 = 2k \equiv 0\pmod 2$ number of pairs, in the end the sequence of letters must look like \[LLLLL\dots RRRRR\]where $L$ appears $\ell_1$ times and $R$ (say) $\ell_2$ times with \[\ell_1+\ell_2 = 2k\] The crucial observation now is that the $L$'s alternate giving $-2, +2$ (to the score) from the left end, whereas the $R$'s alternate giving $-2, +2$ from the right end as well. When $\ell_1, \ell_2$ are both even, the contributions clearly add up to $0$ and the game ends in a draw. On the other hand, if $\ell_1, \ell_2$ are both odd, the contributions add up to $-4$, implying that Anya has won. Thus the above strategy indeed prevents Becky from winning. This shows that Becky cannot always ensure a win for $n$ divisible by $4$. To summarise what we've shown so far, we note that for all $n \not \equiv 2 \pmod 4$, Becky cannot always ensure a win. Thus, to solve the problem, it suffices to prove that she can always win when $n \equiv 2 \pmod 4$. Let the numbers written down be \[a_1, a_2, \dots, a_{2k}\]where $n = 2k$ with $k$ odd. Suppose for the sake of contradiction, that Becky does not have a winning strategy. Write a $0$ or $1$ at each position, depending on whether the index is even or odd. We then have \[101010\dots10\]where $1,0$ each appear $k$ times. We show that Becky can always ensure that Anya only picks out elements marked $0$. We let Becky start by removing the $1$ on the left end, leading to a string with both ends $0$; forcing Anya to pick out some element marked $0$. Once she does, Becky can instantly choose the $1$ that opens up (on either end) and force Anya back into a string with both ends $0$. This can be repeated until the string is empty, at which point Anya has picked all the $0$ elements, and Becky has picked all the $1$ elements. Note that by reversing the initial string, Becky could also have equivalently chosen all the $0$ elements, and Anya all the $1$ elements. Now consider $S_1 = \sum \limits_{i \text{ even}} a_i$ and $S_2 = \sum \limits_{i \text{odd}} a_i$. If $S_1 \ne S_2$, from the previous paragraph we see that Becky can force a game with score $|S_1-S_2| > 0$, thereby winning. Thus we must have \[S_1 = S_2\]This implies that \[2 \mid S_1 + S_2 = 1+2+\dots + 2k = k(2k+1)\]which is impossible since $k, 2k+1$ are both odd. This contradiction shows that there is no game with $n \equiv 2 \pmod 4$ that Becky cannot win. To conclude, the required $n$ are precisely the ones that satisfy $n \equiv 2 \pmod 4$.