This is Problem 1 from Japan MO Finals 2024. See here.

Tintarn wrote: This is Problem 1 from Japan MO Finals 2024. See here. Indeed, though here I give the version with complex numbers. Of course, the AoPS post contains a solution to the complex version as well, but during the competition there were people who knew about the original formulation of the Japanese problem (real numbers) and took them several hours to realize how to transfer to complex numbers

Woah, this is cool. Note that we take indices modulo $n$. Choose the $a_i$ with maximum modulus. Then we have $|a_{i -1} - 2a_i| \ge 2|a_{i}| - |a_{i-1}| \ge |a_i|$, which implies equality everywhere. In particular, this gives us $|a_{i-1} = |a_i|$, and therefore all the numbers have same modulus. Assume this modulus is nonzero, since otherwise we have the trivial solution $(0, \cdots, 0)$. Now, assume that for some $i$ we have $a_{i-1} \ne a_i$. We turn to the complex plane. Note that $a_{i-1} - 2a_i$ lies on the same line as $a_{i-1}, a_i$, further, it must lie on the circle $|z| = |a_1| = r$. But a circle may intersect a line in at most two points, so three of $a_{i-1}, a_i,a_{i-1} - 2a_i$ are equal. Now $a_{i-1} - 2a_i = a_i \implies a_{i-1} = 3a_i \implies r = 3r$, contradiction. If $a_{i-1} - 2a_i = a_{i-1}$, then $a_i = 0$, giving the trivial solution again. So we must have $a_i = a_{i-1}$. Repeating this argument over all $i$, we get all $a_i$ are equal to some $a$. But now $a-2a = a \implies a = 0$, yielding the trivial solution yet again. So the only solution is $\boxed{(a_1, \cdots, a_n) = (0, \cdots, 0)}$.