ニ、(Full score 25 points) Write 1995 equations of the form $A_{i} x^{2}+B_{i} x+C_{i}=0(i=1,2, \cdots, 1995)$ on the blackboard. Two people, A and B, take turns to play a game. Each time, only one non-zero real number is allowed to replace one of the $A_{i}, B$ or $C$ in the $n$ equations $(i=1,2, \cdots, 1995)$. Once a coefficient in the equation is filled, it cannot be changed. After $3 \times 1995$ steps, 1995 quadratic equations with specific real coefficients are obtained. A always tries to make as many quadratic equations as possible have no roots, while B always tries to prevent A from achieving his goal. If A fills first, how many rootless equations can A get at most, regardless of what B does?
Second, the answer is that Jia can get at most 998 rootless equations.
If Jia first fills in the coefficient of the linear term $B_{1}$ with 1, and tries to fill in the linear term coefficient of each equation with 1 as much as possible, at this time, Jia has at least $\frac{1995+1}{2}=998$ chances.

If Yi fills in the coefficient of the $x^{2}$ term (or the constant term) with $a$ in the equations where Jia has already filled in the coefficient of the linear term, then Jia will fill in the third position with $\frac{1}{a}$, resulting in the equation $a x^{2}+x+\frac{1}{a}=0$, or $\frac{1}{a} x^{2}+x+a=0$. At this point, $\Delta=1-4 \cdot \frac{1}{a} \cdot a=-3<0$, so the equation has no roots. Therefore, Jia can get at least 998 rootless equations.