1. Natural numbers from 1 to 2023 are written in a row in some order. It turned out that any three numbers, located every other one, give different remainders when divided by 3. Which number can be in the first place?
Answer: any number that gives a remainder of 1 when divided by 3.

Solution. Let's divide the range from 1 to 2023 into groups of numbers that give remainders of 0, 1, and 2 when divided by 3. The first and third groups contain 674 numbers each, while the second group contains 675 numbers. In any sequence of six consecutive numbers, there are exactly two numbers from each group, since the positions $1,3,5$, as well as the positions $2,4,6$, contain numbers from different groups. We will divide the range from the second to the 2023rd position into blocks of 6 numbers. Each block contains the same number of numbers from each group, and thus the entire range does as well. The remaining number gives a remainder of 1 when divided by 3, and only it can be the first.

We will show that any number from the second group can be in the first position. The feasibility of the configuration of numbers depends not on the numbers themselves, but only on their remainders when divided by 3. Let's arrange these remainders according to the scheme

$$
1220011220011 \ldots 220011
$$

This scheme satisfies the condition of the problem. Therefore, any number that gives a remainder of 1 when divided by 3 can be the first.