17. Find the sum of the reciprocals of all positive divisors of 360.
17. Let the sum of all positive divisors of a positive integer $a$ be $b$. Let $d_{1}$, $d_{2}, \cdots, d_{n}$ be the positive divisors of $a$ arranged in ascending order, so $d_{1}=1, d_{n}=a$. Since the least common multiple of the denominators in $S=\sum_{i=1}^{n} \frac{1}{d_{i}}$ is $d_{n}=a$, we have
$$
S=\sum_{i=1}^{n} \frac{d_{n+1-i}}{d_{n}}=\frac{1}{d_{n}} \sum_{i=1}^{n} d_{i}=\frac{b}{a} .
$$

Given $a=360=2^{3} \times 3^{2} \times 5$, we have
$$
b=\left(1+2+2^{2}+2^{3}\right) \times\left(1+3+3^{2}\right) \times(1+5)=1170 \text {. }
$$

Therefore, the sum of the reciprocals of all positive divisors of 360 is $\frac{1170}{360}=\frac{13}{4}$.