1. Numbers

Suppose the positive integer $n$ is odd. First Al writes the numbers $1,2, \ldots, 2n$ on the blackboard. Then he picks any two numbers $a,b$ erases them and writes instead $|a-b|$. Prove that an odd number will remain at the end.

2. Think I got it. I had to look for the invariant.

3. In the beginning there are $n$ even integers there are $n$ odd integers.

When we replace two the parity of the odd integers stays odd at each step. So we have 2 integers left 1 is odd and 1 is even. And when we subtract them we have an odd integer.

4. I said the sum initially is $n(2n+1)$ which is odd. The sum is reduced by $2 \text{min}(a,b)$ each time which is even. So the parity stays constant, and the sum is $1\mod 2$. The parity is odd at the beginning. So it is odd at the end.

5. Originally Posted by tukeywilliams
I said the sum initially is $n(2n+1)$ which is odd. The sum is reduced by $2 \text{min}(a,b)$ each time which is even. So the parity stays constant, and the sum is $1\mod 2$. The parity is odd at the beginning. So it is odd at the end.
It is similar to mine. But I find your approach more elegant.