Solution:
Answer: $108$, $117$, $135$, $180$. Sum $540 = 108 \times 5 = 135 \times 4 = 180 \times 3$.

Try looking for a number of the form $3n$, $4n$, $5n$, $n$. We want $12n$, $15n$ and $20n$ to have the same first digit. If the first digit is $1$, this requires $n = 9$. We must now check that the fourth number which must be $60n - 12n - 15n - 20n = 13n$ also has three digits starting with $1$. It does, so we are home. [In fact, in this case the first digit must be $1$, since $20n > \frac{3}{2} 12n$.]