A Complex Function

I'll start off by claiming that any odd positive integer $\leq 50$ is attainable and thus the sum of the solutions will simply be (using the formula for the sum of consecutive odd numbers) $25^2=\boxed{625}$ . Here's why:

The initial sum of the numbers on the board is simply $1+2+3+4\dots=1275$ , which is odd. Each operation of replacing $a$ and $b$ with $|a-b|$ (assuming WLOG $a \leq b$ ) will decrease the sum by $2a$ :

$S{\text{new}}=S{\text{old}-|a-b+|a-b}=S{\text{old}}-b-a+b-a=S{\text{old}}-2a$ .

This immediately implies that the new sum must be odd if the old sum was odd, and therefore no even number can result from repeated applications of such an operation starting with $1275$ .

Furthermore, all the numbers on the board are always nonnegative. They're also all less than or equal to $50$ .

We will now show that any odd integer from $1$ to $49$ , inclusive, can be obtained by applying the puzzles's operation $49$ times. Let $k$ be such a number. We can obtain it by subtracting $1$ from $k+1$ . Then we can apply the operation to the pairs of the remaining consecutive integers

$(2,3),(4,5),\dots, (k-1,k),(k+2,k+3),\dots,(49, 50)$

to get $24$ ones on the board while erasing the above pairs. Applying the operation to the $24$ pairs of ones yields $12$ zeros which can be reduced to a single $0$ . Finally, after applying the operation to the remaining numbers, $k+1$ and $1$ , yields $k$ . :D