Sheifer $P$.

The numbers from 1 to 50 are written on the board. It is allowed to erase any two numbers and write down one number which is the absolute value of their difference. After repeating this procedure 49 times, one number will remain on the board. What can this number be?
Clearly, any number written on the board will be between 0 and 50. Moreover, it is odd (see the solution of problem $\underline{30303}$).

We will show that any of the odd numbers from 1 to 49 can be obtained. Suppose we want to get the number $2 m+1$ $(m=0,1, \ldots, 24)$. We divide the numbers from 1 to 50 into pairs as follows: $(1,2 m+2),(2,3),(4,5), \ldots,(2 m, 2 m+1),(2 m+3,2 m$ $+4), \ldots,(49,50)$. We write the absolute difference of the numbers in each pair: $2 m+1,1, \ldots, 1$ (24 ones). It remains to get rid of the ones. For this, we can pair them to get 12 zeros, and then eliminate the zeros (for the pair $(a, 0)$, we can write the single number $|a-0|=a$).

## Answer

Any of the 25 numbers $1,3,5, \ldots, 49$.