Numbers $\frac{49}{1}, \frac{49}{2}, ... , \frac{49}{97}$ are writen on a blackboard. Each time, we can replace two numbers (like $a, b$) with $2ab-a-b+1$. After $96$ times doing that prenominate action, one number will be left on the board. Find all the possible values fot that number.
1. **Define the initial set of numbers:**
   The numbers on the blackboard are \(\frac{49}{1}, \frac{49}{2}, \ldots, \frac{49}{97}\).

2. **Define the operation:**
   Each time, we replace two numbers \(a\) and \(b\) with \(2ab - a - b + 1\).

3. **Identify the invariant:**
   We need to find an invariant that remains unchanged after each operation. Let's consider the product \(\Pi = \prod_{i=1}^n (2a_i - 1)\), where \(a_i\) are the numbers on the blackboard.

4. **Calculate the initial product:**
   Initially, we have:
   \[
   \Pi_{97} = \prod_{i=1}^{97} \left(2 \cdot \frac{49}{i} - 1\right)
   \]
   Simplifying each term:
   \[
   2 \cdot \frac{49}{i} - 1 = \frac{98}{i} - 1 = \frac{98 - i}{i}
   \]
   Therefore:
   \[
   \Pi_{97} = \prod_{i=1}^{97} \frac{98 - i}{i}
   \]
   Notice that the product of the numerators and the product of the denominators are the same:
   \[
   \Pi_{97} = \frac{97!}{97!} = 1
   \]

5. **Verify the invariant under the operation:**
   When we replace \(a\) and \(b\) with \(2ab - a - b + 1\), we need to check if the product \(\Pi\) remains unchanged. Consider:
   \[
   2(2ab - a - b + 1) - 1 = (2a - 1)(2b - 1)
   \]
   This shows that the product of the terms \((2a - 1)\) and \((2b - 1)\) is preserved under the operation.

6. **Conclude the final value:**
   Since the product \(\Pi\) remains invariant and initially \(\Pi_{97} = 1\), the final product when only one number is left must also be 1. Therefore, the last number \(x\) must satisfy:
   \[
   2x - 1 = 1 \implies x = 1
   \]

The final answer is \(\boxed{1}\).