On a blackboard , the 2016 numbers $\frac{1}{2016} , \frac{2}{2016} ,... \frac{2016}{2016}$ are written.
One can perfurm the following operation : Choose any numbers in the blackboard, say $a$ and$ b$ and replace them by $2ab-a-b+1$.
After doing 2015 operation , there will only be one number $t$ Onthe blackboard .
Find all possible values of $ t$.
1. **Initial Setup**: We start with the numbers \(\frac{1}{2016}, \frac{2}{2016}, \ldots, \frac{2016}{2016}\) on the blackboard.

2. **Operation Definition**: The operation allows us to choose any two numbers \(a\) and \(b\) and replace them with \(2ab - a - b + 1\).

3. **Claim**: We claim that \(\frac{1}{2}\) is a fixed point of the operation. To verify this, we substitute \(a = \frac{1}{2}\) into the operation formula:
   \[
   2 \left(\frac{1}{2}\right) b - \left(\frac{1}{2}\right) - b + 1 = b - \frac{1}{2} - b + 1 = \frac{1}{2}
   \]
   This shows that if one of the numbers is \(\frac{1}{2}\), the result of the operation will always be \(\frac{1}{2}\).

4. **Inductive Argument**: We can use induction to show that \(\frac{1}{2}\) will always be present on the blackboard after each operation:
   - Initially, \(\frac{1008}{2016} = \frac{1}{2}\) is on the blackboard.
   - Suppose after \(k\) operations, \(\frac{1}{2}\) is still on the blackboard.
   - In the \((k+1)\)-th operation, we can always choose \(\frac{1}{2}\) and any other number \(b\). The result will be \(\frac{1}{2}\), ensuring that \(\frac{1}{2}\) remains on the blackboard.

5. **Conclusion**: After 2015 operations, only one number remains on the blackboard. Since \(\frac{1}{2}\) is invariant under the operation and is always present, the final number must be \(\frac{1}{2}\).

\(\blacksquare\)

The final answer is \(\boxed{\frac{1}{2}}\)