We wrote the reciprocals of the natural numbers from 2 to 2011 on a board. In one step, we erase two numbers, $x$ and $y$, and write the number

$$
\frac{x y}{x y+(1-x)(1-y)}
$$

in their place. Repeating this 2009 times, only one number remains. What could this number be?
Solution. If

$$
z=\frac{x y}{x y+(1-x)(1-y)}
$$

then

$$
\left(\frac{1}{x}-1\right)\left(\frac{1}{y}-1\right)=\frac{(1-x)(1-y)}{x y}=\left(\frac{1}{z}-1\right)
$$

From this, it follows that if the set of numbers on the board is denoted by $A$, then the value of the product

$$
\prod_{a \in A}\left(\frac{1}{a}-1\right)
$$

does not change during the process. Initially, this value is 2010!, so if the number $v$ remains on the board at the end of the process, then

$$
\frac{1}{v}-1=2010!, \quad \text { i.e., } \quad v=\frac{1}{2010!+1}
$$