1. Answer: $2019^{2021} \cdot 2021^{2019}<2020^{2020} \cdot 2020^{2020}$.
Solution. Let's consider the quotient of these expressions:

$\frac{2019^{2021} \cdot 2021^{2019}}{2020^{2020} \cdot 2020^{2020}}=\frac{2019^{2020} \cdot 2021^{2020}}{2020^{2020} \cdot 2020^{2020}} \cdot \frac{2019}{2020}=\left(\frac{2019 \cdot 2021}{2020^{2}}\right)^{2020} \cdot \frac{2019}{2020}$.

Notice that the last expression has the form $\left(\frac{(n-1)(n+1)}{n^{2}}\right)^{n} \cdot \frac{n-1}{n}$, where each of the factors $\frac{n^{2}-1}{n^{2}}$ and $\frac{n-1}{n}$ is positive but less than 1 for natural $n$. Therefore, the product of these factors is positive but less than 1. This means that the considered quotient is less than 1, so $2019^{2021} \cdot 2021^{2019}$ is justified, but the solution can be fully correct after minor corrections or additions. |
| $3-4$ | Ideas for the solution are provided, the representation as a difference or ratio of fractions, but the transformations contain significant errors or are not completed. |
| $1-2$ | The correct answer is given, supported by some arguments, but the solution lacks a clear justification. |
| 0 | The solution is incorrect, and there is no progress. |
| 0 | The solution is missing. |