On a board, there are $n$ equations in the form $*x^2+*x+*$. Two people play a game where they take turns. During a turn, you are aloud to change a star into a number not equal to zero. After $3n$ moves, there will be $n$ quadratic equations. The first player is trying to make more of the equations not have real roots, while the second player is trying to do the opposite. What is the maximum number of equations that the first player can create without real roots no matter how the second player acts?
To solve this problem, we need to analyze the strategies of both players and determine the maximum number of quadratic equations that the first player can ensure do not have real roots, regardless of the second player's actions.

1. **Understanding the Quadratic Equation**:
   A quadratic equation is of the form \( ax^2 + bx + c = 0 \). The roots of this equation are real if and only if the discriminant \( \Delta = b^2 - 4ac \) is non-negative. Therefore, for the equation to have no real roots, we need \( \Delta < 0 \).

2. **Game Setup**:
   - There are \( n \) quadratic equations.
   - Each equation has three coefficients \( a, b, \) and \( c \) that need to be filled.
   - Players take turns filling in these coefficients with non-zero numbers.
   - After \( 3n \) moves, all coefficients are filled.

3. **Strategy Analysis**:
   - The first player aims to make as many equations as possible have no real roots.
   - The second player aims to make as many equations as possible have real roots.

4. **Case 1: \( n \) is even**:
   - If \( n \) is even, the second player can always counter the first player's moves to ensure that each equation has real roots.
   - For any equation, if the first player fills in \( b \), the second player can choose \( a \) and \( c \) such that \( b^2 - 4ac \geq 0 \).
   - Similarly, if the first player fills in \( a \) or \( c \), the second player can choose the remaining coefficients to ensure real roots.
   - Therefore, the first player cannot guarantee any equation without real roots.

5. **Case 2: \( n \) is odd**:
   - If \( n \) is odd, the first player can use a strategy to ensure that at least \( \frac{n+1}{2} \) equations have no real roots.
   - The first player starts by filling in \( b_1 \) with any number.
   - The second player then fills in another coefficient, say \( a_i \).
   - The first player can then fill in \( c_i \) such that \( b_i^2 - 4a_ic_i < 0 \).
   - By playing symmetrically, the first player can ensure that at least \( \frac{n+1}{2} \) equations have no real roots.

6. **Proof for \( n \) odd**:
   - Lower Bound: The first player can ensure that at least \( \frac{n+1}{2} \) equations have no real roots by strategically filling in the coefficients.
   - Upper Bound: The second player can ensure that at most \( \frac{n-1}{2} \) equations have real roots by countering the first player's moves.

Conclusion:
- For \( n \) even, the first player cannot guarantee any equation without real roots.
- For \( n \) odd, the first player can guarantee that at least \( \frac{n+1}{2} \) equations have no real roots.

The final answer is \( \boxed{ 0 } \) for \( n \) even and \( \frac{n+1}{2} \) for \( n \) odd.