By doing this process, the parity of the sum is invariant because after each operation, the sum diminishes by twice the smaller number taken for the process. So, the initial sum is even implies that the final number(the final sum) is even as well.

Let $N$ be the sums of the squares of the remaining numbers on the blackboard. Originally $N$ is even. The transformation $x,y\to |x-y|$ never affects the parity of $N$ as the difference is $2xy$ (from $x^2+y^2+\ldots$ to $x^2-2xy+y^2\ldots$). So the square of the last number is even, and so is even itself.

Obel1x wrote: A little boy wrote the numbers $1,2,\cdots,2011$ on a blackboard Do you really think that a little boy can do that ?

Thalesmaster wrote: Obel1x wrote: A little boy wrote the numbers $1,2,\cdots,2011$ on a blackboard Do you really think that a little boy can do that ? In a recent British MO, it said there were $2010^{2010}$ students at a camp! Sometimes maths problems neglect a sense of realism, I think

The sum of the numbers on the board is invariant and originally even. The result follows directly

Pascal96 wrote: The sum of the numbers on the board is invariant and originally even. The result follows directly modulo $2$

Obel1x wrote: A little boy wrote the numbers $1,2,\cdots,2011$ on a blackboard. He picks any two numbers $x,y$, erases them with a sponge and writes the number $|x-y|$. This process continues until only one number is left. Prove that the number left is even. If the problem asks "What could be the last number on the board?", are all even numbers achievable?

dattranquoc101201 wrote: If the problem asks "What could be the last number on the board?", are all even numbers achievable? Obviously all even numbers aren't achievable.Because the processes are finite or the sum of the numbers decrease after each operation.

Let $S=1+2+3+...+2011$ $S= \frac{2011 \cdot 2012}{2} = 2011 \cdot 1006 = 2023066$, which is even Consider 3 cases: $x>y \Rightarrow S-x-y+(x-y)=S-2y$, which is even $x<y \Rightarrow S-x-y+(y-x)=S-2x$, which is even $x=y \Rightarrow S-2x+(x-x)=S-2x$, which is even Therefore the last number on the board is always even $\blacksquare$