NZMO Round Two

Solution:
Notice that $r_{n}(x)$ is just $x$ modulo $n$. Therefore
$$\prod_{i} r_{n}(i a_{i}) \equiv \prod_{i} i a_{i} \pmod{n}$$
For primes $p$, apply Wilson's theorem to see that we must have
$$\prod_{i} i a_{i} \equiv -1 \pmod{p}$$
However,
$$\prod_{i} i a_{i} = \prod_{i} i \prod_{i} a_{i} \equiv (-1)^{2} \equiv 1 \pmod{p}$$
So the only possibility in that case is $p = 2$.

Now, if $n$ is composite, then let $n$ be the minimal solution. Let $n = pq$ for prime $p$. Notice that we must have $(p - 1)q$ numbers in $\{a_{1}, 2a_{2}, 3a_{3}, \ldots , (n - 1)a_{n - 1}\}$ not divisible by $p$. The $q - 1$ numbers of the form $(kp)a_{kp}$ are divisible by $p$ and there are only $q - 1$ multiples of $p$ in $\{1, 2, \ldots , pq - 1\}$. Therefore they must be the only numbers divisible by $p$, so $\{a_{kp} \mid 1 \leq k < q\} = \{kp \mid 1 \leq k < q\}$.

Now, $p \nmid q$ as otherwise all of $(kp)a_{kp}$ are multiples of $p^{2}$, which is not true.

Let $c \equiv \frac{1}{p} \pmod{q}$. Consider $\left\{\frac{a_{p}}{p}, \frac{a_{2p}}{p}, \ldots , \frac{a_{(q - 1)p}}{p}\right\} = \{1, 2, \ldots , q - 1\}$.
Then,
$$\left\{\frac{a_{p}}{p}, \frac{2a_{2p}}{p}, \ldots , \frac{(q - 1)a_{(q - 1)p}}{p}\right\} = \left\{\frac{p a_{p}}{p^{2}}, \frac{2p a_{2p}}{p^{2}}, \ldots , \frac{(q - 1)p a_{(q - 1)p}}{p^{2}}\right\}$$
$$\equiv c\{1,2,3,\ldots ,q - 1\} \pmod{q}$$
As $(k p)a_{k p} \equiv k' p$ (mod $p q$) for some $k'$, and $k a_{k p} \equiv k'$ (mod $q$)
Since $\gcd(c,q) = 1$
$$c\{1,2,3,\ldots ,q - 1\} \equiv \{1,2,3,\ldots ,q - 1\} \pmod{q}$$
in some order. This is a solution for $n = q$, which contradicts the minimality of the solution for $n = p q$. Therefore no such solution $n = p q$ can exist.

Hence, $n = 2$ is the only solution.