The numbers $1,\frac12,\frac13,\ldots,\frac1{1999}$ are written on a blackboard. In every step, we choose two of them, say $a$ and $b$, erase them, and write the number $ab+a+b$ instead. This step is repeated until only one number remains. Can the last remaining number be equal to $2000$?
1. **Initial Observation**:
   We start with the numbers \(1, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{1999}\) on the blackboard. In each step, we choose two numbers \(a\) and \(b\), erase them, and write the number \(ab + a + b\) instead.

2. **Transformation Analysis**:
   Let's analyze the transformation \(ab + a + b\). Notice that:
   \[
   ab + a + b + 1 = (a + 1)(b + 1)
   \]
   This implies that the transformation \(ab + a + b\) can be rewritten as:
   \[
   ab + a + b = (a + 1)(b + 1) - 1
   \]

3. **Invariant Quantity**:
   Consider the product of all numbers on the board increased by 1:
   \[
   (a_1 + 1)(a_2 + 1)(a_3 + 1) \cdots (a_k + 1)
   \]
   Initially, this product is:
   \[
   (1 + 1)\left(\frac{1}{2} + 1\right)\left(\frac{1}{3} + 1\right) \cdots \left(\frac{1}{1999} + 1\right)
   \]

4. **Initial Product Calculation**:
   Calculate the initial product:
   \[
   (1 + 1)\left(\frac{1}{2} + 1\right)\left(\frac{1}{3} + 1\right) \cdots \left(\frac{1}{1999} + 1\right)
   \]
   Simplify each term:
   \[
   2 \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{2000}{1999}
   \]
   Notice that this is a telescoping product, where most terms cancel out:
   \[
   2 \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{2000}{1999} = 2000
   \]

5. **Final Step**:
   Since the product \((a_1 + 1)(a_2 + 1)(a_3 + 1) \cdots (a_k + 1)\) remains constant throughout the process, and initially it is 2000, the final remaining number \(x\) must satisfy:
   \[
   x + 1 = 2000 \implies x = 1999
   \]

6. **Conclusion**:
   Therefore, the last remaining number on the board can indeed be 2000.

The final answer is \( \boxed{ 2000 } \)