The numbers $1,2,...,n$ are written on a board ($n\in\mathbb N$). In each step we take any two numbers $a,b$, remove them, and write either $a-b$ or $a+b$ on the board. After $n-1$ steps there will be one number $t$ on the board. What numbers $t$ are achievable after $n-1$ steps?
It's clear that the parity of the sum of the numbers is invariant. So the parity of $t$ must match the parity of $1+\cdots+n=n(n+1)/2$. Can all the parity matching numbers be achieved? Thanks for any help.
It never hurts to gather some numeric data:
$$\begin{array}{c|l} n&\text{possible results}\\ \hline 1&1\\ 2&-1,1,3\\ 3&-4,-2,0,2,4,6\\ 4&-8,-6,-4,-2,0,2,4,6,8,10\\ 5&-13,-11,-9,-7,-5,-3,-1,1,3,5,7,9,11,13,15 \end{array}$$
At this point it’s not hard to conjecture what the minimum and maximum possible outcomes are in terms of $n$, and that every integer of the same parity lying between those values is also a possible outcome. You can then prove the conjecture by induction on $n$. As Michael suggested in the comments, in the induction step consider first what happens when you leave the largest number for last. Then show that no other order of play can produce a result outside that range.