Count!

If the sum of three integers is odd, then two cases are possible:

1. All three numbers are odd.
2. Two numbers are even, and the third number is odd.

To pick three distinct odd numbers from the given set, there are $\dbinom{5}{3} = \dfrac{5!}{2! \times 3!} = 10$ ways.

To pick two even and one odd number from the set, there are $\dbinom{5}{2} \times \dbinom{5}{1} = 50$ ways.

Therefore, the three numbers can be chosen in $10 + 50 = \boxed{60}$ ways. $_\square$