1. For how many natural numbers $n$ are the inequalities

$$
2019<\sqrt{n}<2020 ?
$$

Explain your answer!
Solution. By squaring the given inequalities, we obtain the equivalent inequalities $2019^{2}<n<2020^{2}$. According to this, the total number of numbers for which the last inequalities, i.e., the given inequalities, are true is equal to

$$
2020^{2}-2019^{2}-1=(2020-2019)(2020+2019)-1=4039-1=4038
$$