For any 4 distinct points $P_{1}, P_{2}, P_{3}, P_{4}$ in the plane, find the minimum value of the ratio $\frac{\sum_{1 \leq i<j \leq 4} P_{i} P_{j}}{\min _{1 \leq i<j \leq 4} P_{i} P_{j}}$.
First, we provide a lemma.

Lemma: In $\triangle ABC$, $BC \geqslant \min \{AB, AC\} \cdot 2 \sin \frac{A}{2}$.

Proof of the lemma: As shown in Figure 23-5, draw the angle bisector $AT$ of $\angle A$, and drop perpendiculars from $B$ and $C$ to $AT$, with the feet of the perpendiculars being $D$ and $E$ respectively. Then,
$$
\begin{aligned}
BC & = BT + TC \geqslant BD + CE = (AB + AC) \sin \frac{A}{2} \\
& \geqslant \min \{AB, AC\} \cdot 2 \sin \frac{A}{2}.
\end{aligned}
$$

Thus, the lemma is proved.

Solution to the original problem: Let $m = \min_{1 \leqslant i < j \leqslant 4} P_i P_j$, and $S = \sum_{1 \leqslant i < j \leqslant 4} P_i P_j$. Given $\alpha + \beta > 60^\circ$, we have $\frac{\alpha + \beta}{4} \geqslant 45^\circ$ and $0^\circ \leqslant \frac{\alpha - \beta}{4} < 15^\circ$.

By the lemma, $P_2 P_4 \geqslant 2m \sin \frac{\alpha}{2}$ and $P_1 P_3 \geqslant 2m \sin \frac{\beta}{2}$, so
$$
P_1 P_3 + P_2 P_4 \geqslant 2m \left( \sin \frac{\alpha}{2} + \sin \frac{\beta}{2} \right)
$$
$$
\begin{array}{l}
= 4m \sin \frac{\alpha + \beta}{4} \cos \frac{\alpha - \beta}{4} \\
\geqslant 4m \cdot \sin 45^\circ \cdot \cos 15^\circ = (\sqrt{3} + 1)m.
\end{array}
$$

At this point, we also have $\frac{S}{m} \geqslant 5 + \sqrt{3}$.
When $P_1 P_2 P_3 P_4$ is a rhombus with internal angles of $60^\circ$, $\frac{S}{m} = 5 + \sqrt{3}$.
In summary, the minimum value of $\frac{S}{m}$ is $5 + \sqrt{3}$.