1. In some cells of a $1 \times 2100$ strip, one chip is placed. In each of the empty cells, a number is written that is equal to the absolute difference between the number of chips to the left and to the right of this cell. 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: 1400.

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

$$
2100-n \leqslant\left[\frac{n+1}{2}\right] \leqslant \frac{n+1}{2}, \quad \text { hence } n \geqslant \frac{4199}{3}=1399 \frac{2}{3} \text { and } n \geqslant 1400
$$

We will show that the value $n=1400$ is achievable. The arrangement

$$
\underbrace{0101 \ldots 01}_{700 \text { pairs }} \underbrace{111 \ldots 1}_{700 \text { numbers }}
$$

where 1s represent chips and 0s represent empty cells, will work. In this case, the numbers in the positions of the zeros will be:

$$
1400,1398,1396, \ldots, 4,2
$$

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

$$
\underbrace{101101 \ldots 101}_{350 \text { triples }} \underbrace{110110 \ldots 110}_{350 \text { triples }}
$$

The corresponding numbers are:

$$
1398,1394, \ldots, 6,2,4,8, \ldots, 1396,1400
$$