## Task B-4.2.

On the planet "Anything is Possible," children play a game in which each of them has to choose numbers from the set of odd natural numbers less than 1000 within a certain time. In this game, the sum of no pair of different numbers chosen by any child is 1002. The winner is the one who selects the most such numbers. How many numbers can the winner select at most? At the end of the game, it was found that several children had selected the maximum number of such numbers, and no two children had selected all the same numbers. Determine the maximum number of children for whom this could have happened.
Solution.

There are 500 odd numbers less than 1000. We group them into pairs whose sum is 1002:

$$
(3,999),(5,997), \ldots,(497,505),(499,503)
$$

There are 249 such pairs. Both numbers in a pair cannot be chosen. The numbers 1 and 501 do not have a pair, so they can be chosen.

We conclude that a child, or the winner, can choose at most 251 numbers, i.e., one number from each of the pairs $(*)$ and the numbers 1 and 501.

Let's determine the maximum number of children who could each choose 251 numbers, with at least one chosen number being different for each of them.

All of them must have chosen the numbers 1 and 501. From each of the pairs $(*)$, one number can be chosen in two ways. Since there are 249 pairs, the number of possible selections is $2^{249}$. Therefore, the maximum number of children who could each choose 251 numbers with the given property is $2^{249}$.