Game with 2020! consecutive integers, each turn select a and b, then write a new number on the board, which is a + b − ab/100 
On the board, there are 2020! consecutive integers, from 1 to 2020!. Kevin plays the following
  game: Each turn, he erases two random numbers, assume that they are a and b, then he writes a new number
  on the board, which is a + b −
  ab/100 . After 2020! − 1 turns, there is one number left on the board. Does that
  number depend on what number Kevin erases?(If not, what is that number?(Prove it.)If yes, prove it.)

I don't know how to approach this problem.
This problem comes from COFFEE contest problem 4 (it already ended).

Solution 1:
Just note that, if one of $a$ and $b$ is $100$, then the number $a + b - ab/100$ remains $100$.
In other words, the number $100$ "eats" any other number.
Thus in whatever order he erases and writes numbers, in the end the number $100$ will eat all other numbers and thus the final number must be $100$.

Solution 2:
As a more general approach, for every number $x$, let $f(x)$ denote $1 - x / 100$.
It is then easy to see that $f(a)f(b) = f(a + b - ab/100)$.
This means that however he erases and writes numbers, the product of $f(x)$ for all numbers $x$ on the board does not change.
Since there is originally $100$ on the board, and $f(100)$ is zero, the final number $x$ must also satisfy $f(x) = 0$ and hence $x = 100$.
