The numbers $\frac{1}{1}, \frac{1}{2}, \cdots , \frac{1}{2012}$ are written on the blackboard. Aïcha chooses any two numbers from the blackboard, say $x$ and $y$, erases them and she writes instead the number $x + y + xy$. She continues to do this until only one number is left on the board. What are the possible values of the final number?
1. Let's denote the numbers on the blackboard as \( a_1 = \frac{1}{1}, a_2 = \frac{1}{2}, \ldots, a_{2012} = \frac{1}{2012} \).

2. Aïcha chooses any two numbers \( x \) and \( y \) from the blackboard, erases them, and writes the number \( x + y + xy \). We can rewrite this operation as:
   \[
   x + y + xy = (1 + x)(1 + y) - 1
   \]

3. Suppose Aïcha first chooses \( a_1 \) and \( a_2 \). The new number she writes is:
   \[
   (1 + a_1)(1 + a_2) - 1
   \]

4. Next, she chooses this new number and \( a_3 \). The result of this operation is:
   \[
   \left[(1 + a_1)(1 + a_2) - 1 + 1\right](1 + a_3) - 1 = (1 + a_1)(1 + a_2)(1 + a_3) - 1
   \]

5. By continuing this process, we can generalize that after \( n \) steps, the number on the board will be:
   \[
   \left[\prod_{k=1}^{n} (1 + a_k)\right] - 1
   \]

6. When only one number is left on the board, it will be:
   \[
   \left[\prod_{k=1}^{2012} (1 + a_k)\right] - 1
   \]

7. Now, we need to calculate the product:
   \[
   \prod_{k=1}^{2012} (1 + a_k) = \prod_{k=1}^{2012} \left(1 + \frac{1}{k}\right)
   \]

8. Notice that:
   \[
   1 + \frac{1}{k} = \frac{k + 1}{k}
   \]

9. Therefore, the product becomes:
   \[
   \prod_{k=1}^{2012} \frac{k + 1}{k} = \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{2013}{2012}
   \]

10. This is a telescoping product, where all intermediate terms cancel out, leaving:
    \[
    \frac{2013}{1} = 2013
    \]

11. Thus, the final number on the board is:
    \[
    2013 - 1 = 2012
    \]

The final answer is \(\boxed{2012}\).