The numbers $\frac{1}{1}, \frac{1}{2}, ... , \frac{1}{2010}$ are written on a blackboard.  A student chooses any two of the numbers, say $x$, $y$, erases them and then writes down $x + y + xy$. He continues to do this until only one number is left on the blackboard. What is this number?
1. **Initial Setup**: We start with the numbers $\frac{1}{1}, \frac{1}{2}, \ldots, \frac{1}{2010}$ on the blackboard.

2. **Transformation Rule**: The student chooses any two numbers, say $x$ and $y$, erases them, and writes down $x + y + xy$. This can be rewritten using the identity:
   \[
   x + y + xy = (x+1)(y+1) - 1
   \]
   This identity is crucial because it helps us track the product of the terms in a simplified form.

3. **Invariant Product**: Define $z_i = x_i + 1$ for each number $x_i$ on the blackboard. Initially, we have:
   \[
   z_1 = 1 + \frac{1}{1} = 2, \quad z_2 = 1 + \frac{1}{2} = \frac{3}{2}, \quad \ldots, \quad z_{2010} = 1 + \frac{1}{2010} = \frac{2011}{2010}
   \]
   The product of all $z_i$'s is:
   \[
   \prod_{i=1}^{2010} z_i = \prod_{i=1}^{2010} \left(1 + \frac{1}{i}\right)
   \]

4. **Product Simplification**: The product can be simplified as follows:
   \[
   \prod_{i=1}^{2010} \left(1 + \frac{1}{i}\right) = \prod_{i=1}^{2010} \frac{i+1}{i} = \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{2011}{2010}
   \]
   Notice that this is a telescoping product, where most terms cancel out:
   \[
   \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{2011}{2010} = \frac{2011}{1} = 2011
   \]

5. **Final Number**: Since the product of all $z_i$'s remains invariant and equals 2011, the final number on the blackboard, when only one number is left, must satisfy:
   \[
   z_{\text{final}} = 2011
   \]
   Therefore, the final number $x_{\text{final}}$ is:
   \[
   x_{\text{final}} + 1 = 2011 \implies x_{\text{final}} = 2010
   \]

The final answer is $\boxed{2010}$