The following figure shows a  $4\times 5$  array of points, each of which is 1 unit away from its nearest neighbours. The number of non-degenerate triangles (i.e., triangles with positive area whose vertices are points in the given array is divisible by 

There are $\left(\begin{array}{c}20\\ 3\end{array}\right)=\frac{20\cdot 19\cdot 18}{3!}=1140$  ways to choose three points. But not all of these triplets form non-degenerate triangles. We need to find how many triplets are collinear. We consider the five cases shown in following figures.
            
In case (i), there are four lines, each containing five points. Hence there are  $4\times \left(\begin{array}{c}5\\ 3\end{array}\right)=40$  collinear triplets. In case (ii) and case (iii) combined, there are nine lines altogether, each containing four points. Hence there are  $9\times \left(\begin{array}{c}4\\ 3\end{array}\right)=36$  collinear triplets. In case (iv) and case (v) combined, there are eight lines altogether, each containing three points. Thus there are  $8\times \left(\begin{array}{c}3\\ 3\end{array}\right)=8$ collinear triplets. Thus the answer is  $1140-(40+36+8)=1056$.