IMO Hk TST

The answer is $37$.

We first prove the following result.
Claim. Consider a positive rational number $\frac{a}{b}$ where $(10a, b) = 1$. The smallest period of this number when expressed as a decimal number is the order of $10$ modulo $b$.
Proof. Note that $\frac{a}{b}$ has period $k$ if and only if the tail part of $10^k \cdot \frac{a}{b}$ is the same as that of $\frac{a}{b}$. This holds if and only if $(10^k - 1) \cdot \frac{a}{b}$ is a terminating decimal, i.e. $10^m(10^k - 1) \frac{a}{b} \in \mathbb{Z}$ for some nonnegative integer $m$. Equivalently, this means
$$b \mid 10^m(10^k - 1)a.$$
As $(10a, b) = 1$, this is the same as $b \mid 10^k - 1$. By definition, the smallest positive integer $k$ for which this holds is the order of $10$ modulo $b$. $\square$

Now, when $n$ runs through all positive integers, the denominator of $\frac{n}{567}$ in the lowest term can be any positive divisor of $567 = 3^4 \times 7$. Let $d_b$ be the order of $10$ modulo $b$. By the claim, it suffices to find $d_b$ for each positive divisor $b$ of $567$.
* Since $10 \equiv 1 \pmod{9}$, we have $d_1 = d_3 = d_9 = 1$.
* Since $10 \not\equiv 1 \pmod{27}$ and $10^3 = 1000 \equiv 1 \pmod{27}$, we have $d_{27} = 3$.
* Since $10^3 = 1000 \not\equiv 1 \pmod{81}$ and
$$10^9 = 1000^3 = (27 \times 37 + 1)^3 \equiv 1 \pmod{81}$$
by the binomial theorem, we have $d_{81} = 9$.
* Since $\varphi(7) = 6$ and $10^2, 10^3 \not\equiv 1 \pmod{7}$, we have $d_7 = 6$.
* For $b = 3^s7$ where $s = 1, 2, 3, 4$, we have $d_b = [d_{3^s}, d_7]$. Therefore, we easily deduce $d_{21} = d_{63} = d_{189} = 6$, $d_{567} = 18$.
It follows that the answer is $1 + 3 + 6 + 9 + 18 = 37$.