$2021 \cdot 1009 = 2039189$.
Let's solve the problem for an arbitrary odd number $n$ of marked points and drawn segments. We enumerate the marked points on the circle by the numbers from $1$ to $n$ in the order of going around the circle. For each $i = 1, \overline\{n\}$ we denote by $a_i$ the number of drawn segments for which $i$ is an endpoint. Among all $\binom\{n\}\{2\}$ pairs of segments, at least $\binom\{a_i\}\{2\}$ pairs do not intersect, since they have a common endpoint $i$. For different $i = 1, \overline\{n\}$ these pairs do not coincide, hence the number of intersection points does not exceed
$$M = \binom\{n\}\{2\} - \left( \binom\{a_1\}\{2\} + \binom\{a_2\}\{2\} + \dots + \binom\{a_n\}\{2\} \right).$$
Let's determine the largest possible value of $M$, for which we find the smallest possible value of the sum in brackets. It is easy to see, by opening the brackets, that the inequality $\binom\{b\}\{2\} + \binom\{a\}\{2\} > \binom\{b-1\}\{2\} + \binom\{a+1\}\{2\}$ is equivalent to the inequality $b-1 > a$. Therefore, the numbers $a_i, i = 1, \overline\{n\}$, must differ at most by one. Given that $a_1 + a_2 + \dots + a_n = 2n$, it's only possible if $a_1 = a_2 = \dots = a_n = 2$. Hence, the maximum possible value of $M$ is equal to $\frac\{n(n-1)\}\{2\} - n = \frac\{n(n-3)\}\{2\}$.
To make sure that this value is the answer for the problem, we provide an example. Let's arrange points at the vertices of a regular $n$-gon and sequentially perform the following operation with each of them: choose a point $i$ and consider the set $p_i$ of intersection points of all possible segments with vertices at the remaining points. The set $p_i$ is finite, and the set of points on the arc containing the point $i$ is infinite. Therefore, we can choose such a new location for it that no segment connecting the point $i$ with any other point passes through the points of the set $p_i$. As a result of such operations, we obtain an arrangement of points such that no three segments with vertices at these points have a common point inside the circle. Now it suffices to draw segments connecting the points with numbers $i$ and $i + [n/2]$, $i = 1, \overline\{n\}$, (we assume that the numbers are cyclic modulo $n$, i.e. $n + k = k$ if $k = 1, \overline\{n\}$). In this case, any two segments that don't have a common endpoint, intersect and all $a_i$ are equal to two.
Substituting the value $n = 2021$ into the found formula $\frac\{n(n-3)\}\{2\}$, we find the answer $2021 \cdot 1009 = 2039189$.