Numbers from $1,\frac12,\frac13,...........\frac{1}{2010}$ are written and any two $x,y$ are taken and we replace $x,y$ by just $x+y+xy$

This is an invariant question : imagine a function $f(x_1,...,x_m)$ (where $m$ is a certain number of arguments and $x_i$ are all real numbers) with the following property : $f(x_1,...,x_m)$ doesn't change when you take any two of the these $x_i,x_j$ and replace them by just $x_i+x_j+x_ix_j$.

Then what happens? If there's just one number $N$ on the board left after all that, then $f(x_1,...,x_m) = f(N)$, so $N = f^{-1}(f(x_1,...,x_m))$ provided that $f(x_1,...,x_m)$ has exactly one preimage.

A hint for this function $f$ comes from $(1+x)(1+y)=1+(x+y+xy)$, so something like : add $1$ to all the numbers you have, and multiply these results together?

It is obvious that such a function does the job! In which case, we must add $1$ to each of the numbers, and multiply them all. That's like multiplying $\frac{2}{1}, \frac 32, \frac 43 ,...\frac {2011}{2010}$, which is just $2011$.

Now, whatever last number is on the board, one plus that is $2011$, so it is $2010$.

The operation $x*y=x+y+xy=(x+1)(y+1)-1$ on real numbers is associative so the result does not depend on the order of steps and is equal to $$(1+1)(1+1/2)...(1+1/2010)-1=2011!/2010!-1=2010$$

Suppose you choose $\frac1m$ and $\frac1n$ in the first turn, replace them by $\left(\frac{m+1}m\frac{n+1}n\right)-1$

(note that $x+y+xy=(x+1)(y+1)-1$)

In the next turn, you might choose two numbers $\frac1a$ and $\frac1b$, and the replaced number will look just as above, with $a,b$ replacing $m,n$. However, if you choose the new number obtained in the previous step, i.e. $\left(\frac{m+1}m\frac{n+1}n\right)-1$ and one of the original numbers $\frac1a$, then you replace them by $\left(\frac{m+1}m\frac{n+1}n\frac{a+1}a\right)-1$.

Fill in the intermediate steps to show by induction that the replaced number in any step will look like $\left(\prod_j\frac{a_j+1}{a_j}-1\right)$, so that the final answer will be $$\dfrac{2011}{2010}\dfrac{2010}{2009}\cdots \dfrac{2}{1}-1=2010$$.