For any division of the 2021 points into 20 groups $M_1, M_2, \dots, M_{20}$ of different size we let $n_k$ be the number of points in the group $M_k$ and we suppose that $n_1 < n_2 < n_3 < \dots < n_{20}$. The number $g = n_{20} - n_1 - 19$ is then equal to the number of integers between $n_1$ and $n_{20}$ which are not equal to any $n_k$.

The main observation is that when $g \ge 2$ it is possible to increase the number of triangles by moving one point to another group. To see this, let $n_i < n_j$ be such that $n_i + 1 < n_j - 1$ and the two numbers $n_i + 1$ and $n_j - 1$ are not among the $n_k$. Such $i, j$ always exist when $g \ge 2$, because we can pick two of the missing integers to be $n_i + 1$ and $n_j - 1$.

When we move a point $P$ from $M_j$ to the group $M_i$, we gain triangles with one vertex $P$ and another vertex in $M_j \setminus \{P\}$. If $r = 2021 - n_i - n_j$ is the number of point in the 18 unaffected groups, then the number of triangles gained is equal to $r(n_j - 1)$. But we also lose all the triangles that had $P$ as one vertex and another vertex in $M_i$. The number of lost triangles is equal to $rn_i$. Hence, the net gain when we move $P$ from $M_j$ to $M_i$ is
$$r(n_j - 1) - rn_i = r(n_j - n_i - 1) > 0.$$
Because the number of triangles is bounded above by $\binom{2021}{3}$, we cannot increase the number of triangles indefinitely. Hence, starting with any division of the 2021 points into 20 groups of different size and moving points between groups in the way described above, we will reach a situation at which an increase of the number of triangles is not possible. In such a situation we must have $g = 0$ or $g = 1$. This means that the numbers $n_1, n_2, \dots, n_{20}$ are obtained by removing one number from a list of 21 consecutive numbers. If the smallest or largest number is removed from the list we have $g = 0$, and $g = 1$ otherwise.

Let $n_1$ be the smallest number of this list of 21 consecutive numbers and let $n_1 + k$ be the missing number, where $1 \le k \le 20$. We then have
$$2021 = \sum_{i=1}^{20} n_i = \sum_{i=0}^{20} (n_1 + i) - (n_1 + k) = 20n_1 + 210 - k.$$
This gives $1811 = 20n_1 - k$ and so $k = 9$ and $n_1 = 91$. Therefore, $g = 1$ and the numbers $n_1, n_2, \dots, n_{20}$ are the integers from 91 to 111, except 100. If these are the sizes of the groups, the maximum number of triangles is achieved. In any other case, the number of triangles is below the maximum.