Four, (50 points) In the 21st Century City, all streets run either east-west or north-south. To enhance security management, security booths are placed at some intersections (no two security booths are on the same street), and the rectangle formed by two security booths as two vertices and streets as edges is called a security zone. The number of security booths within a security zone (including the boundary) is called the security strength of that zone.

If the city has at least $n$ $(n \geqslant 5)$ streets in both directions, and any two non-parallel streets intersect at an intersection, now select $n$ intersections to place security booths according to the requirements. Find the minimum value of the maximum security strength of any security zone.
Let the southernmost, easternmost, northernmost, and westernmost security booths be $A, B, C, D$ (some may coincide). The east-west streets passing through $A, C$ and the north-south streets passing through $B, D$ form a rectangle $M$, within which all security booths are located. Let $n=5k+r (0 \leqslant r \leqslant 4, k \geqslant 1)$, and use $S(X)$ to denote the security strength of the security zone $X$.
(1) If at least two different points among $A, B, C, D$ are vertices of $M$, then $M$ itself is a security zone. In this case,
$$
S(M)=n \geqslant\left[\frac{n}{5}\right]+2.
$$
(2) If exactly one of $A, B, C, D$ is a vertex of $M$ (let it be $A$). In this case, each of the two sides of $M$ not containing $A$ has one security booth (let them be $B, C$, as shown in Figure 4). Then the three security zones $AB, BC, CA$ cover $M$. Thus, the $n-3$ security booths outside $A, B, C$ are covered by the above three security zones. Therefore, at least one security zone $X$ covers at least $\left[\frac{5k-3}{3}+\frac{2}{3}\right] \geqslant k$ of these $n-3$ security booths. Since $X$ covers two points among $A, B, C$ (with two of these points as vertices), we have
$$
S(X) \geqslant k+2=\left[\frac{n}{5}\right]+2.
$$
(3) If none of $A, B, C, D$ are vertices of $M$, then the four security zones $AB, BC, CD, DA$ cover all security booths in $M$ except for the rectangle $A'B'C'D'$ (as shown in Figure 5). The security zone $AC$ covers the rectangle $A'B'C'D'$. Thus, the $n-4$ security booths outside $A, B, C, D$ are covered by the above five security zones. Therefore, at least one security zone $Y$ covers at least $\left[\frac{5k-4}{5}\right]+1=k$ of these $n-4$ security booths. Since $Y$ covers two points among $A, B, C, D$ (with two of these points as vertices), we have
$$
S(Y) \geqslant k+2=\left[\frac{n}{5}\right]+2.
$$

From the above, we see that $S_{\max} \geqslant\left[\frac{n}{5}\right]+2$.
Next, divide the $n=5k+r$ security booths into five groups, with the number and distribution of security booths in each group as shown in Figure 6, where $r$ groups on the boundary contain $k+1$ security booths, and the other groups contain $k$ security booths.
For any two security booths $P, Q$:
When $P, Q$ belong to the same group, $S(PQ) \leqslant k+1$.
When exactly one of $P, Q$ belongs to the central group, the security zone $PQ$ either contains exactly one point from the central group or exactly one point from a non-central group, so
$$
S(PQ) \leqslant 1+(k+1)=k+2.
$$

When $P, Q$ belong to two adjacent boundary groups, the security zone $PQ$ either contains exactly one point from one of these groups or exactly one point from the other group, so
$$
S(PQ) \leqslant 1+(k+1)=k+2.
$$

When $P, Q$ belong to two opposite boundary groups, the security zone $PQ$ contains exactly one point from each of these groups and at most $k$ points from the central group, so
$$
S(PQ) \leqslant 1+k+1=k+2.
$$

Furthermore, in Figure 6, there clearly exist security booths $P, Q$ such that $S(PQ)=k+2$, thus $S_{\text{max}}=k+2$.
$$
\text{Therefore, }\left(S_{\max}\right)_{\min}=\left[\frac{n}{5}\right]+2.
$$