4. In the Cartesian coordinate system, mark all integer points $(x, y)(0 \leqslant y \leqslant 10)$. Question: For a polynomial of degree 20 with integer coefficients, what is the maximum number of marked integer points that can lie on this polynomial?
4. For a polynomial $P(x)$ with integer coefficients of degree 20, we only need to consider the number of integer solutions to the inequality $0 \leqslant P(x) \leqslant 10$.

Prove by contradiction that there are no more than 20 solutions, otherwise, there would be at least 21.

Let these solutions that satisfy the inequality, in ascending order, be $x_{1}, x_{2}, \cdots, x_{21}$. Define the polynomial
$$
Q(x)=P(x)-10 \text {. }
$$

Thus, when $x=x_{2}, x_{3}, \cdots, x_{21}$, $Q(x)=0$. This indicates that the 18-degree polynomial $Q(x)$ has more than 18 roots, which is a contradiction.
On the other hand,
$$
P(x)=\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{20}\right)
$$

has 20 integer solutions $x_{1}, x_{2}, \cdots, x_{20}$ that satisfy $0 \leqslant P(x) \leqslant 10$.

Therefore, the conclusion of this problem is that there are at most 20 points marked on the graph of a 20-degree polynomial with integer coefficients.