1. We are given 20 points with no three collinear and need to determine how many obtuse triangles can be formed by these points.
2. To solve this, we need to understand the conditions under which a triangle is obtuse. A triangle is obtuse if one of its angles is greater than 90 degrees.
3. Consider the construction of points such that every set of three points forms an obtuse triangle. We can construct the points from left to right, labeling them $A_1, A_2, \ldots, A_{20}$.
4. For points $A_m$ and $A_n$ with $m > n$, denote the perpendicular at $A_m$ to $A_mA_n$ by $P_{(m,n)}$. Denote the set of points to the right of $P_{(m,n)}$ by $R_{(m,n)}$.
5. $A_3$ must lie in $R_{(1,2)}$. Then $A_4$ must lie in the intersection of $R_{(1,2)}$, $R_{(1,3)}$, and $R_{(2,3)}$. Continue in this way for the remaining points: $A_k$ must lie in the intersection of $R_{(m,n)}$ ranging over all pairs $(m,n)$ with $1 \le n < m < k$.
6. It is clear that you can choose points such that this region is always nonempty. Therefore, every set of three points forms an obtuse triangle.
7. The number of ways to choose 3 points from 20 is given by the binomial coefficient $\binom{20}{3}$.

$$\binom{20}{3} = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$$

8. Therefore, the number of obtuse triangles that can be formed is $\binom{20}{3}$.

The final answer is $\boxed{\binom{20}{3}}$.