4. 2 .

Let $f(x)=a(x-s)^{2}+t$.
If $|t| \leqslant 50$, then when $|x-s| \geqslant 1$,
$$
f(x)>100(x-s)^{2}+t \geqslant 100+t \geqslant 50 \text {. }
$$

Therefore, when $|x-s|50$, we discuss in two cases.
(1) If $t>50$, then $f(x)>t>50$ always holds. Hence, there are no integers that satisfy the requirement.
(2) If $t<-50, s$ is the axis of symmetry of the quadratic function $f(x)$. We will prove that there is at most one integer on one side of the axis of symmetry that satisfies $|f(x)| \leqslant 50$. Otherwise, suppose
$$
s<n<n+k\left(n \in \mathbf{Z}, k \in \mathbf{Z}_{+}\right)
$$

and $|f(n)| \leqslant 50,|f(n+k)| \leqslant 50$.
Then $|f(n+k)-f(n)| \leqslant 100$
$$
\Leftrightarrow a\left|2 n k-2 s k+k^{2}\right| \leqslant 100 \text {. }
$$

Thus, $|k(2 n-2 s+k)|<1$, which is clearly a contradiction.
Hence, there is at most one integer on one side of the axis of symmetry that satisfies $|f(x)| \leqslant 50$.
Therefore, at most 2 integers satisfy the requirement.
In particular, taking $f(x)=101\left(x-\frac{1}{2}\right)^{2}(x=0,1)$ satisfies the requirement.
4. Let $f(x)=a x^{2}+b x+c$, where, $a, b, c \in$ $\mathbf{R}, a>100$. Then the number of integer $x$ that satisfies $|f(x)| \leqslant 50$ is at most $\qquad$.

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