1. In some cells of a $1 \times 2021$ strip, one chip is placed in each. For each empty cell, the number equal to the absolute difference between the number of chips to the left and to the right of this cell is written. It is known that all the written numbers are distinct and non-zero. What is the minimum number of chips that can be placed in the cells?
Answer: 1347.

Solution. Let $n$ be the number of chips placed. Note that the numbers in the empty cells lie in the range from 1 to $n$ and have the same parity. Therefore, there can be no more than $\left[\frac{n+1}{2}\right]$ such numbers. This means that the number of empty cells does not exceed $\left[\frac{n+1}{2}\right]$, otherwise the numbers placed in them would repeat. Then

$$
2021-n \leqslant\left[\frac{n+1}{2}\right] \leqslant \frac{n+1}{2}, \quad \text { hence } \quad n \geqslant \frac{4041}{3}=1347
$$

We will show that the value $n=1347$ is achievable. A suitable arrangement is

$$
\underbrace{0101 \ldots 01}_{674 \text { pairs }} \underbrace{111 \ldots 1}_{673 \text { numbers }}
$$

where ones represent chips, and zeros represent empty cells. In this case, the numbers in the positions of the zeros will be:

$$
1347,1345,1343, \ldots, 3,1
$$

Remark. The implementation provided in the solution is not unique. For example, the following also works:

$$
1011011 \ldots 0110
$$

The corresponding numbers are:

$$
1345,1341, \ldots, 5,1,3,7, \ldots, 1343,1347
$$