2. Find four such numbers that all their pairwise sums are consecutive natural numbers, the smallest of which is 2023.
Solution. Note that the pairwise sums of the numbers $1, 3, 5$, and $9$ give six consecutive even numbers from $4$ to $14$. Then the pairwise sums of the numbers $n+1, n+3, n+5, n+9$ also form six consecutive even numbers from $2n+4$ to $2n+14$. Dividing them by 2, we get six consecutive natural numbers. Therefore, the set of numbers $\frac{n+1}{2}$, $\frac{n+3}{2}$, $\frac{n+5}{2}$, $\frac{n+9}{2}$ works. Given that $\frac{n+1}{2} + \frac{n+3}{2} = 2023$, we have $n = 2021$, and we get the set $1011, 1012, 1013, 1015$. Answer: $1011, 1012, 1013, 1015$.

Comment. Without any explanation, the correct example is worth 7 points.