, 2024

Let $f(a, b) = \text{gcd}(2024ab, a^2 + 254ab + b^2)$ then for 2 numbers $a, b$ on the board, the new number generated will be $f(a, b)$. We consider some cases:
* If $a, b$ are same parity then $f(a, b)$ is even.
* If $a, b$ are different from parity then $f(a, b)$ is odd.
This means that the number of odd numbers either remains the same or decreases by 2 units; but initially, there are odd numbers, so the final number $x$ must be odd. Suppose that $x > 1$, let $p$ be some prime divisor of $x$.
Note that
$$254 = 11 \cdot 23 + 1 \text{ and } 2024 = 2^3 \cdot 11 \cdot 23.$$
If $p = 11$, since $11|\text{gcd}(2024ab, a^2 + 254ab + b^2)$ then $11|a^2 + ab + b^2$. Similarly, if $p = 23$ then there is also $23|a^2 + ab + b^2$. We have the following familiar lemma derived from Fermat's little theorem or quadratic residue:
Lemma. If $p$ is a prime of form $3k + 2$ and $p|a^2 + ab + b^2$ then $p|a, b$.
Since 11 has the form $3k + 2$, according to the above lemma, one can get $11|a, b$. Continuing like that, the previous numbers on the board that generated $a, b$ also be divisible by 11, making all the original numbers divisible by 11, contradiction. Similarly for the case $p = 23$.
Finally, if $p \neq 11, 23$ then $p|ab, p|a^2 + 506ab + b^2$ entails $p|ab, p|a^2 + b^2$. Obviously from $p|ab$, we get that $p$ is a divisor of one of $a$ or $b$, but $p|a^2 + b^2$ then $p|a, b$. We get the similar contradiction as the above argument.
Therefore $x$ is odd and has no odd prime divisors, proving that $x = 1$. ☐