Figure 2

A point $M(k,l)$ will belong in the interior of the line segment $OA_i$, if and only if $\overrightarrow{OM}$ and $\overrightarrow{OA_i}$ have the same slope (with $k,l$ integers and $0 < k \le 40$), that is
$$\frac{i}{40} = \frac{l}{k} \quad (\text{with } k,l \text{ integers and } 0 < k \le 40).$$
In order the line segment $OA_i$ be "good" it is enough and sufficient the fraction $\frac{i}{40}$ to be reducible (then we have fraction $\frac{l}{k}$ with integer terms equivalent to $\frac{i}{40}$ and its terms give the coordinates of the "good" point $M(k,l)$).
Hence, if $(40,i) > 1$, then the line segment $OA_i$ is "good" and we have $(40,i)-1$ "good" points at the segment $OA_i$. To the point $A_2(40,2)$ corresponds the "good" segment $OA_2$, which contains the good point $(20,1)$. To point $A_4(40,4)$ corresponds the "good" segment $OA_4$, which contain the "good" points $(10,1)$, $(20,2)$, $(30,3)$. Working in this way we finally find 24 "good" segments and 140 "good" points.

An easy solution can be given by using the Euler function. $\phi$. It is known that the number of positive integers lower or equal to $n$ and not relatively prime to $n$ is $n-\phi(n)$. Since $40 = 5 \cdot 2^3$, we have
$$\phi(40) = 40 \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{5}\right) = 40 \frac{14}{25} = 16.$$
Hence the number of good segments is $40 - \phi(40) = 24$.