1. Albert makes a sequence of numbers, the first 2023 of which are exactly the numbers 1 through 2023, in some order. To determine each subsequent number, Albert takes the median of the previous 2023 numbers. You find the median of 2023 numbers by sorting them by size and taking the exact middle number from that sequence.

How many different values can the three-thousandth number in Albert's sequence have?
B1. 1 Note that the median of 2023 numbers is always the number that stands at position 1012 when you sort the numbers by size. The first number that Albert adds, and which thus ends up at position 2024, is the median of the numbers from 1 to 2023. That is 1012. We will now show that all subsequent numbers that Albert adds are also equal to 1012.

Suppose that Albert has just written the number 1012 at position $n$, as the median of the numbers at positions $n-2023$ to $n-1$. We call this the old set. Which number will appear at position $n+1$? For this, Albert looks at the set of numbers at positions $n-2022$ to $n$, which we call the new set. The new set is almost the same as the old set: the number 1012 has been added and the number at position $n-2023$ is no longer included. Let's call this last number $x$.

We now consider the three options for $x$. If $x=1012$, then the new set consists of exactly the same numbers as the old set, and thus the median is again 1012. If $x$ is greater than 1012, then the smallest 1012 numbers from the old set remain the same (the median of the old set is 1012, after all) and the median is again 1012. If $x$ is less than 1012, then the largest 1012 numbers in the new set are exactly the largest 1012 numbers from the old set, and in this case the median is also equal to 1012.

We conclude that Albert writes down the number 1012 every time. Therefore, the three-thousandth number is also equal to 1012. There is thus only one possibility.