Answer: 100.

Solution:

In each equation, there must be a coefficient less than five hundred. Otherwise, the discriminant $b^{2}-4 a c<1000^{2}-4 \cdot 500^{2}=0$ and the equation has no roots. Thus, we get no more than 100 equations.

Now for an example: Take $b$ as numbers starting with 9, and for each equation, take $a$ and $c$ as $500-k$ and $399+k$ respectively, where $k$ takes natural values from 0 to 99. Then $a c=(500-k)(399+k)<(500-k)(400+k)=200000+100 k-k^{2}=202500-(k-50)^{2} \leqslant 202500$. Therefore, $4 a c<810000$, on the other hand, $b^{2} \geq 900^{2}=810000$, hence $b \geqslant a c$.
8. (4 points) On the board, all three-digit natural numbers are written, the first digits of which are odd and greater than 1. What is the maximum number of quadratic equations of the form $a x^{2}+b x+c=0$ that can be formed using these numbers as $a, b$, and $c$, each no more than once, such that all these equations have roots.