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 $S_{\text{max}}=k+2$.
$$\text{Therefore, }\left(S_{\max}\right)_{\min}=\left[\frac{n}{5}\right]+2.$$