jmc

In the worst-case scenario, every possible sum is rolled before the same sum is rolled again. The minimum possible sum rolled is $3\cdot 1=3$, and the maximum is $3\cdot 6=18$. Every sum in between those two extremes can be created, since the sums are created through adding three of the digits between one and six. Thus, there are $18-2=16$ possible sums, so the dice must be rolled $\boxed{17}$ times to ensure that the same sum is rolled twice.