The answer is all odd $n$. Even $n$ evidently do not work, since we may take the $n$ numbers $1,2,1,2,\ldots,1,2$ arranged in a circle. Now, for odd $n$, let $a_i$ be our $n$ numbers, and note that the condition rewrites as $a_{i+2}-a_i=a_{i+3}-a_{i+1}$ for all $n$, where indices are taken $\pmod n$. Note that $a_3-a_1=a_4-a_2=a_5-a_3=\ldots=a_n-a_{n-2}=a_1-a_{n-1}=a_2-a_n,$ and if we let $k$ be that common difference, we infer that $a_1+\ldots+a_n=(a_3+a_3+\ldots+a_n+a_1+a_2)+kn,$ and so $k=0$. Therefore, $a_1=a_3=a_5=\ldots=a_n=a_2=a_4=a_6=\ldots=a_{n-1},$ and so all $n$ numbers are equal, as desired.