There are a number of cards on a table. A number is written on each card. The "pick and replace" operation involves the following: two random cards are taken from the table and replaced by one new card. If the numbers $a$ and $b$ appear on the two packed cards, the number $a + b + ab$ is set on the new card.
If we start with ten cards with the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ and $10$ respectively, what value(s) can the number have that "grab and replace" nine times is on the only card still on the table? Prove your answer
1. **Initial Setup**: We start with ten cards numbered \(1, 2, 3, 4, 5, 6, 7, 8, 9, 10\).

2. **Operation Definition**: The "pick and replace" operation involves selecting two cards with numbers \(a\) and \(b\), and replacing them with a new card with the number \(a + b + ab\).

3. **Transformation**: Notice that the operation \(a + b + ab\) can be rewritten as:
   \[
   a + b + ab = (a + 1)(b + 1) - 1
   \]
   This transformation will be useful in simplifying the problem.

4. **First Operation**: Suppose we pick the first two cards with numbers \(a_1\) and \(a_2\). The new number on the card will be:
   \[
   b_1 = a_1 + a_2 + a_1a_2 = (a_1 + 1)(a_2 + 1) - 1
   \]

5. **Second Operation**: Next, we pick the new card with number \(b_1\) and another card with number \(a_3\). The new number will be:
   \[
   b_2 = b_1 + a_3 + b_1a_3 = ((a_1 + 1)(a_2 + 1) - 1 + 1)(a_3 + 1) - 1 = (a_1 + 1)(a_2 + 1)(a_3 + 1) - 1
   \]

6. **General Operation**: After the \(k\)-th operation, where \(1 \leq k \leq 9\), the new number \(b_k\) can be expressed as:
   \[
   b_k = (a_1 + 1)(a_2 + 1) \cdots (a_{k+1} + 1) - 1
   \]

7. **Final Operation**: After the 9th operation, we will have combined all ten numbers into one final number. This final number \(b_9\) will be:
   \[
   b_9 = (a_1 + 1)(a_2 + 1) \cdots (a_{10} + 1) - 1
   \]

8. **Substitution**: Substituting the initial numbers \(1, 2, 3, \ldots, 10\) into the expression, we get:
   \[
   b_9 = (1 + 1)(2 + 1)(3 + 1) \cdots (10 + 1) - 1 = 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \cdot 11 - 1
   \]

9. **Simplification**: The product \(2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8 \cdot 9 \cdot 10 \cdot 11\) is \(11!\) (11 factorial). Therefore:
   \[
   b_9 = 11! - 1
   \]

Conclusion:
\[
\boxed{11! - 1}
\]