Numbers $1,\frac12,\frac13,\ldots,\frac1{2001}$ are written on a blackboard. A student erases two numbers $x,y$ and writes down the number $x+y+xy$ instead. Determine the number that will be written on the board after $2000$ such operations.
1. **Understanding the operation:**
   The operation described in the problem is to erase two numbers \( x \) and \( y \) and write down the number \( x + y + xy \). This can be rewritten using the identity:
   \[
   x + y + xy = (x+1)(y+1) - 1
   \]
   This identity will be useful in simplifying the problem.

2. **Transforming the problem:**
   Let's consider the transformation \( z = x + 1 \) and \( w = y + 1 \). Then the operation becomes:
   \[
   (x+1)(y+1) - 1 = zw - 1
   \]
   This means that instead of working with \( x \) and \( y \), we can work with \( z \) and \( w \) where \( z = x+1 \) and \( w = y+1 \).

3. **Initial transformation:**
   The initial numbers on the blackboard are \( 1, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{2001} \). Transforming these numbers using \( z = x + 1 \), we get:
   \[
   2, \frac{3}{2}, \frac{4}{3}, \ldots, \frac{2002}{2001}
   \]

4. **Product of transformed numbers:**
   The product of these transformed numbers is:
   \[
   2 \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \ldots \cdot \frac{2002}{2001}
   \]
   Notice that this is a telescoping product, where most terms cancel out:
   \[
   2 \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \ldots \cdot \frac{2002}{2001} = \frac{2002}{1} = 2002
   \]

5. **Final transformation:**
   After 2000 operations, only one number remains on the board. Let this number be \( N \). According to our transformation, this number \( N \) corresponds to:
   \[
   N + 1 = 2002 \implies N = 2001
   \]

Conclusion:
\[
\boxed{2001}
\]