 Navi and Ozna are playing a game where Ozna starts first and the two take turn making moves. A positive integer is written on the waord. A move is to (i) subtract any positive integer at most 2015 from it or (ii) given that the integer on the board is divisible by $2014$, divide by $2014$. The first person to make the integer $0$ wins. To make Navi's condition worse, Ozna gets to pick integers $a$ and $b$, $a\ge 2015$ such that all numbers of the form $an+b$ will not be the starting integer, where $n$ is any positive integer.

Find the minimum number of starting integer where Navi wins.
To solve this problem, we need to determine the minimum number of starting integers where Navi wins, given the rules of the game and the constraints imposed by Ozna.

1. **Define the Winning Function:**
   Let \( f(n) = 1 \) if the first player (Ozna) wins with starting integer \( n \), and \( f(n) = 2 \) if the second player (Navi) wins with starting integer \( n \). We also define \( f(0) = 2 \) since the player who makes the integer 0 wins.

2. **Determine the Winning Conditions:**
   - \( f(k) = 1 \) if there exists an \( x < 2016 \) such that \( f(x) = 2 \), or if \( 2014 \mid x \) and \( f(x/2014) = 2 \).
   - \( f(k) = 2 \) if for all \( x < 2016 \), \( f(x) = 1 \), and if \( 2014 \mid x \), then \( f(x/2014) = 1 \).

3. **Identify the Pattern:**
   - The smallest value where \( f(x) = 2 \) is \( x = 0 \).
   - The next value where \( f(x) = 2 \) is \( x = 2016 \) if \( 2014 \) does not divide \( 2016 \).
   - Continuing this pattern, the values where \( f(x) = 2 \) up to \( 2014 \times 2016 \) are the multiples of \( 2016 \).

4. **Consider the Constraints:**
   - Ozna picks integers \( a \) and \( b \) such that \( a \ge 2015 \) and all numbers of the form \( an + b \) will not be the starting integer.
   - The AP (Arithmetic Progression) that Ozna picks must contain the AP of the form \( 2015^2 + 2016n \) entirely, hence its common difference must divide \( 2016 \).

5. **Determine the Minimum Number of Starting Integers Where Navi Wins:**
   - The AP that Ozna picks must have a common difference of \( 2016 \).
   - The values where Player 2 (Navi) wins are the multiples of \( 2016 \) up to \( 2014 \times 2016 \).
   - Since these values cannot be contained in the AP chosen by Ozna, all \( 2013 \) of them are available.

Thus, the minimum number of starting integers where Navi wins is \( 2013 \).

The final answer is \(\boxed{2013}\).