The only such $N$ are powers of $2$.

If $N$ is not a power of $2$, let's say $p \nmid N$ where $p$ is an odd prime. Now, observe that $p \nmid 1$ but $p \mid 2^k - 1$ for some $k > 1$. Then, let $\alpha = \frac{N}{p}$. Now,
$$a_{\alpha} > a_{2\alpha} \ge a_{4\alpha} \dots \ge a_{2^k \alpha} \implies a_{\alpha} > a_{2^k \alpha}$$
But now, $a_{2^k \alpha} = a_{\alpha}$ as $N|(2^k - 1)\alpha$ but then this is a contradiction! We have that $a_{\alpha} > a_{2\alpha}$ since all reals in the circle are distinct and $N \nmid \alpha$.

Now, if $N = 2^m$ is a power of $2$, let $a_j = -\nu_2(j) + \frac{j}{2N}$. This is clearly okay if $j = N$ and if $j \ne N$ then $\nu_2(2j) = \nu_2(j) + 1$ and thus $a_j \ge a_{2j}$. $\square$