Suppose set $\Omega = \{ r \mid r = \frac{p}{q},\ p, q \in \mathbb{N}_+,\ (p, q) = 1,\ pq \mid 3600 \}$.
We consider the reduced fractional form $\frac{p}{q}$ of any element $r$ of $\Omega$. Since the standard factorization of $3600$ is $2^4 \times 3^2 \times 5^2$, we can set $p = 2^A \times 3^B \times 5^C$, $q = 2^a \times 3^b \times 5^c$, where $\min\{A, a\} = \min\{B, b\} = \min\{C, c\} = 0$ and $A + a \le 4$, $B + b \le 2$, $C + c \le 2$.
Therefore, there are $9$ ways to take such number pair $(A, a)$, $5$ ways to take number pair $(B, b)$, and $5$ ways to take number pair $(C, c)$.
As a result, the number of the elements of $\Omega$ is $|\Omega| = 9 \times 5 \times 5 = 225$.
All the rational numbers satisfying the conditions are $\Omega \cap (0, 1)$. Notice that $r \in \Omega$ if and only if $\frac{1}{r} \in \Omega$. In particular, $1 \in \Omega$. Therefore, the elements in $\Omega \setminus \{1\}$ can be matched into
$$\frac{1}{2}(|\Omega| - 1) = 112$$
pairs according to the product of $1$, and each pair has exactly one number belonging to $(0, 1)$, that is, there is exactly one number satisfying the conditions. Thus, the number of desired rational numbers $r$ is $112$. $\square$