Four. (50 points) Let $n$ be a positive integer, and let the planar point set be
$$
S=\{(x, y) \mid x, y \in\{0,1, \cdots, n\}, x+y \neq 0\} \text {. }
$$

Question: What is the minimum number of lines in the plane whose union can contain $S$, but not include the point $(0,0)$?
Four, at least $2 n$ lines are needed.
It is easy to see that $2 n$ lines satisfy the requirements. For example, the lines $x=i$ and $y=i(i=1,2, \cdots, n)$, it is easy to see that the union of these $2 n$ lines contains $S$ but does not contain the origin.
Another example is the set of lines
$$
x+y+z=k(k=1,2, \cdots, 2 n) \text {. }
$$

Below is the proof that $2 n$ is the smallest possible number.
Assume that the union of $N$ lines in the plane contains $S$ but does not contain the origin, and let their equations be
$$
a_{i} x+b_{i} y+c_{i}=0(1 \leqslant i \leqslant N) .
$$

Consider the polynomial
$$
P(x, y)=\prod_{i=1}^{N}\left(a_{i} x+b_{i} y+c_{i}\right) \text {. }
$$

Then its degree is $N$, and for any $(x, y) \in S$, we have
$$
P(x, y)=0, P(0,0) \neq 0 \text {. }
$$

Let $Q(y)=y(y-1) \cdots(y-n)$, and let $R(x, y)$ be the remainder when $P(x, y)$ is divided by $Q(y)$.

Since the polynomial $Q(y)$ has $n+1$ zeros at $y=0,1, \cdots, n$, we know
$$
P(x, y)=R(x, y)
$$

for all $x, y \in\{0,1, \cdots, n\}$.
Note that $\operatorname{deg} Q(y)=n+1$.
Thus, $\operatorname{deg}_{y} R(x, y) \leqslant n$, and it is clear that
$$
N=\operatorname{deg} P(x, y) \geqslant \operatorname{deg} R(x, y) \text {. }
$$

Express the polynomial $R(x, y)$ in the form of a decreasing power of $y$: $R(x, y)=R_{n}(x) y^{n}+R_{n-1}(x) y^{n-1}+\cdots+R_{0}(x)$.
Since $R(0,0)=P(0,0) \neq 0$, $R(0, y)$
is not the zero polynomial.
Furthermore, when $y=1,2, \cdots, n$, $R(0, y)=P(0, y)=0$.
Therefore, $R(0, y)$ has $n$ roots.
Thus, $R_{n}(0) \neq 0$, indicating that $R_{n}(0)$ is not the zero polynomial.
For $i \in\{1,2, \cdots, n\}$, when $y=0,1, \cdots, n$, we have
$$
R(i, y)=P(i, y)=0 .
$$

This indicates that $R(i, y)$ has at least $n+1$ roots.
Since $\operatorname{deg} R(i, y) \leqslant n$, $R(i, y)$ is the zero polynomial.
Therefore, for any $i \in\{1,2, \cdots, n\}, R_{n}(i)=0$.
Thus, $R_{n}(x)$ has at least $n$ roots.
Since $R_{n}(0)$ is not the zero polynomial, we have
$\operatorname{deg} R(x) \geqslant n$.
Thus, $\operatorname{deg} R(x, y) \geqslant \operatorname{deg} R(x)+n \geqslant 2 n$.
Therefore, $N \geqslant 2 n$.
In summary, at least $2 n$ lines are needed to satisfy the conditions of the problem.
(Math Studio Wenwu Guanghua, 215128)