smc

If the first number of a group of $n$ consecutive numbers is $a$, the $n^{\text{th}}$ number is $a+n-1$. We know that the sum of the group of numbers is $100$, so $$\frac{n(2a+n-1)}{2}=100$$ $$2a+n-1=\frac{200}{n}$$ $$2a=1-n+\frac{200}{n}$$ We know that $n$ and $a$ are positive integers, so we check values of $n$ that are a factor of $200$. Of these values, the only ones that result in a positive integer $a$ is when $n=5$ or when $n=8$, so there are $\boxed{2}$ sets of two or more consecutive positive integers whose sum is $100$.