Solution:

Each element equals $0.\overline{abc} = \frac{100a + 10b + c}{999},$ and there are $10 \cdot 9 \cdot 8 = 720$ ordered triples of distinct digits. By symmetry, each digit $0$ through $9$ appears in each of the three positions exactly $\frac{720}{10} = 72$ times.

The numerators therefore total $72 (0 + 1 + \cdots + 9)(100 + 10 + 1) = 72 \cdot 45 \cdot 111 = 359640,$ so the sum of the elements is $\frac{359640}{999} = 360.$