1. The natural numbers from 1 to 2021 are written in a row in some order. It turns out that for any number, its left and right neighbors have different parity. Which number can be in the first place?
Answer: any odd number.

Solution. First, let's prove that in any sequence of four consecutive numbers, there are two even and two odd numbers. Indeed, suppose in some quartet there are three numbers of the same parity. Then among them, there will be two numbers that are two positions apart. But the number between them will have neighbors of the same parity, which is impossible.

Let's divide the range from the second to the 2021st position into blocks of 4 numbers. By the proven fact, in each block, there are an equal number of even and odd numbers, and thus, in the entire range as well. However, in the interval $[1,2021]$, there are more odd numbers than even numbers. Therefore, the first number must be odd. We will show that any odd number can be placed in the first position. Arrange all the numbers according to the scheme

## O EEOO EEOO ... EEOO,

where E denotes an even number, and $\mathrm{O}$ denotes an odd number. This arrangement satisfies the condition of the problem. In this case, the odd number on the first position can be arbitrary.