9. 18 Given $f(x)=a x^{2}+b x+c$, where $a, b, c$ are real numbers and $a>100$, how many integers $x$ at most can satisfy $|f(x)| \leqslant 50$?
[Solution] At most two. Otherwise, suppose there are three different integers $x_{1}, x_{2}, x_{3}$ satisfying $\left|f\left(x_{i}\right)\right| \leqslant 50, i=1,2,3$.
Let $x_{1}>x_{2} \geqslant-\frac{b}{2 a}$, since $x_{1}, x_{2}$ are integers,
thus
$$
x_{1}+x_{2}+\frac{b}{a} \geqslant 1 \text {. }
$$

From this, we can get
$$
\begin{aligned}
\left|f\left(x_{1}\right)-f\left(x_{2}\right)\right| & =\left|a\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}+\frac{b}{a}\right)\right| \\
& \geqslant a>100,
\end{aligned}
$$

which contradicts $\left|f\left(x_{1}\right)\right|,\left|f\left(x_{2}\right)\right| \leqslant 50$. Similarly, it can be shown that the case where two of $x_{1}, x_{2}, x_{3}$ are less than or equal to $-\frac{b}{2 a}$ also leads to a contradiction.