Solution e17d604e-6c55-45e8-9b77-4e2cddfca4bb

Points: 

If all three integers are positive, their sum is at least $3$.
Otherwise, exactly two integers must be negative.

Write the integers as $-a,-b,c$, where $1\leq a\leq b$.
A positive sum requires $c>a+b$, so $c$ must be the largest absolute
value. Listing the factor triples $a\leq b\leq c$ with $abc=60$ gives
\[
\renewcommand{\arraystretch}{1.15}
\begin{array}{c|r}
(a,b,c) & c-a-b \\
\hline
(1,1,60) & 58 \\
(1,2,30) & 27 \\
(1,3,20) & 16 \\
(1,4,15) & 10 \\
(1,5,12) & 6 \\
(1,6,10) & 3 \\
(2,2,15) & 11 \\
(2,3,10) & 5 \\
(2,5,6) & -1 \\
(3,4,5) & -2
\end{array}
\]
The least positive sum is therefore attained by $-1,-6,10$:
\[
(-1)(-6)(10)=60,
\qquad
-1-6+10=3.
\]

Final answer:
3