The numbers $1,2,...,100$ are written in a board. We are allowed to choose any two numbers from the board $a,b$ to delete them and replace on the board the number $a+b-1$. 
What are the possible numbers u can get after $99$ consecutive operations of these?
1. Let's denote the initial set of numbers on the board as \( S = \{1, 2, 3, \ldots, 100\} \).
2. We are allowed to choose any two numbers \( a \) and \( b \) from the board, delete them, and replace them with \( a + b - 1 \).
3. We need to determine the final number on the board after 99 operations.

To solve this, we need to understand the effect of each operation on the sum of the numbers on the board.

4. Initially, the sum of the numbers on the board is:
   \[
   \sum_{i=1}^{100} i = \frac{100 \cdot 101}{2} = 5050
   \]

5. Each operation involves choosing two numbers \( a \) and \( b \), removing them, and adding \( a + b - 1 \) to the board. The sum of the numbers on the board after this operation changes as follows:
   \[
   \text{New sum} = \text{Old sum} - a - b + (a + b - 1) = \text{Old sum} - 1
   \]
   This means that each operation decreases the sum of the numbers on the board by 1.

6. Since we perform 99 operations, the total decrease in the sum of the numbers on the board is:
   \[
   99 \times 1 = 99
   \]

7. Therefore, the final sum of the numbers on the board after 99 operations is:
   \[
   5050 - 99 = 4951
   \]

8. After 99 operations, there will be exactly one number left on the board. This number must be the final sum of the numbers on the board.

Conclusion:
The only possible number left on the board after 99 operations is \( \boxed{4951} \).