2A. Below are recorded 2020 real numbers. The sum of each number in the sequence and the reciprocal of the next number is equal to 1. The product of all the numbers is equal to 2020. Which numbers are recorded at the first and last position in the sequence?
Solution. Let the sequence be $a_{1}, a_{2}, \ldots, a_{2020}$. We have $a_{1}+\frac{1}{a_{2}}=1$, so $a_{2}=\frac{1}{1-a_{1}}$. Similarly, from $a_{2}+\frac{1}{a_{3}}=1$ it follows that $a_{3}=\frac{1}{1-a_{2}}=\frac{1}{1-\frac{1}{1-a_{1}}}=\frac{1-a_{1}}{-a_{1}}=1-\frac{1}{a_{1}}$. For the fourth term of the sequence, $a_{4}=\frac{1}{1-a_{3}}=\frac{1}{1-\left(1-\frac{1}{a_{1}}\right)}=a_{1}$. Therefore, the terms of the sequence repeat every three, that is, the sequence is

$$
a_{1}, a_{2}=\frac{1}{1-a_{1}}, a_{3}=1-\frac{1}{a_{1}}, a_{4}=a_{1}, a_{5}=\frac{1}{1-a_{1}}, a_{6}=1-\frac{1}{a_{1}}, a_{7}=a_{1}, \ldots
$$

We have $2020=3 \cdot 673+1$, so $a_{2020}=a_{1}$. For the product of the first three numbers, we have $a_{1} \cdot \frac{1}{1-a_{1}} \cdot\left(1-\frac{1}{a_{1}}\right)=-1$. Then

$$
2020=a_{1} a_{2} \ldots a_{2020}=(-1)^{673} a_{2020}=-a_{1}
$$

from which it follows that $a_{2020}=a_{1}=-2020$.