To determine how many of the numbers $1 \cdot 2 \cdot 3$, $2 \cdot 3 \cdot 4$, ..., $2020 \cdot 2021 \cdot 2022$ are divisible by $2020$, we need to analyze the divisibility conditions.

1. Factorize 2020:
$$2020 = 2^2 \cdot 5 \cdot 101$$
Therefore, a number $n(n+1)(n+2)$ must be divisible by $2^2$, $5$, and $101$.

2. **Divisibility by $2^2$**:
- Among any three consecutive integers, at least one is divisible by $4$ (since every fourth number is divisible by $4$).
- Therefore, $n(n+1)(n+2)$ is always divisible by $4$.

3. **Divisibility by $5$**:
- Among any three consecutive integers, at least one is divisible by $5$ (since every fifth number is divisible by $5$).
- Therefore, $n(n+1)(n+2)$ is always divisible by $5$.

4. **Divisibility by $101$**:
- We need to check how many of the numbers $1, 2, 3, \ldots, 2020$ are divisible by $101$.
- The multiples of $101$ in this range are $101, 202, 303, \ldots, 2020$. There are $20$ such multiples.

5. Counting the valid triplets:
- For each multiple of $101$, we need to check how many triplets $n(n+1)(n+2)$ include this multiple.
- We create a table to count the number of triplets for each multiple of $101$:

$$\begin{array}{|c|c|} \hline \text{Multiple of 101} & \text{Number of triplets} \\ \hline 101 & 2 \\ 202 & 1 \\ 303 & 1 \\ 404 & 2 \\ 505 & 2 \\ 606 & 1 \\ 707 & 0 \\ 808 & 1 \\ 909 & 1 \\ 1010 & 2 \\ 1111 & 1 \\ 1212 & 1 \\ 1313 & 0 \\ 1414 & 1 \\ 1515 & 2 \\ 1616 & 2 \\ 1717 & 1 \\ 1818 & 1 \\ 1919 & 2 \\ 2020 & 3 \\ \hline \end{array}$$

6. Summing the counts:
- Summing the numbers in the second column of the table:
$$2 + 1 + 1 + 2 + 2 + 1 + 0 + 1 + 1 + 2 + 1 + 1 + 0 + 1 + 2 + 2 + 1 + 1 + 2 + 3 = 27$$