15 First prove the following lemma: Let $P$ be any one of the 35 points $P_{1}, P_{2}, P_{3}, \cdots, P_{35}$. Then, among the remaining 34 points, at most 6 points are less than 3 units away from $P$.
(Proof by contradiction) Suppose there are 7 points, say $P_{1}, P_{2}, P_{3}, \cdots, P_{7}$ (arranged counterclockwise), that are less than 3 units away from point $P$.

Since $P_{1}, P_{2}, P_{3}, \cdots, P_{35}$ are the 35 vertices of a convex 35-sided polygon, we have
$$\begin{array}{c} \angle P_{1} P P_{2}+\angle P_{2} P P_{3}+\angle P_{3} P P_{4}+\angle P_{4} P P_{5}+ \\ \angle P_{5} P P_{6}+\angle P_{6} P P_{7} \leqslant 180^{\circ} . \end{array}$$

Therefore, among the 6 angles $\angle P_{1} P P_{2}, \angle P_{2} P P_{3}, \angle P_{3} P P_{4}, \angle P_{4} P P_{5}, \angle P_{5} P P_{6}, \angle P_{6} P P_{7}$, at least one angle is no greater than $30^{\circ}$. Without loss of generality, let $\angle P_{1} P P_{2} \leqslant 30^{\circ}$. Let $P P_{1}=x, P P_{2}=y$, then
$$\begin{aligned} P_{1} P_{2}^{2} & =x^{2}+y^{2}-2 x y \cos \angle P_{1} P P_{2} \\ & \leqslant x^{2}+y^{2}-2 x y \cos 30^{\circ} \\ & =x^{2}+y^{2}-\sqrt{3} x y . \end{aligned}$$

By symmetry, assume $x \geqslant y$. Since $\sqrt{3} \leqslant y \leqslant x<3$, we have
$$P_{1} P_{2}^{2}=f(x)=x^{2}+y^{2}-\sqrt{3} x y=\left(x-\frac{\sqrt{3}}{2} y\right)^{2}+\frac{1}{4} y^{2}$$

is an increasing function on the interval $[y, 3)$. Therefore,
$$\begin{aligned} P_{1} P_{2}^{2} & =f(x)<f(3) \\ & =9+y^{2}-3 \sqrt{3} y \\ & =\left(y-\frac{3 \sqrt{3}}{2}\right)^{2}+\frac{9}{4} \\ & \leqslant\left(\sqrt{3}-\frac{3 \sqrt{3}}{2}\right)^{2}+\frac{3}{4} \\ & =3, \end{aligned}$$

Thus, $P_{1} P_{2}<\sqrt{3}$, which contradicts the given condition. Therefore, the assumption is false.
Hence, the lemma is proved.
Now, using the lemma to prove the conclusion of the problem.
According to the lemma, among the 34 segments $P_{1} P_{2}, P_{1} P_{3}, P_{1} P_{4}, \cdots, P_{1} P_{35}$ starting from $P_{1}$, at most 6 segments have a length less than 3, i.e., at least 28 segments have a length not less than 3. Without loss of generality, assume the segments $P_{1} P_{2}, P_{1} P_{3}, P_{1} P_{4}, \cdots, P_{1} P_{29}$ have a length not less than 3.

Next, consider the 27 segments $P_{2} P_{3}, P_{2} P_{4}, P_{2} P_{5}, \cdots, P_{2} P_{29}$ starting from $P_{2}$. According to the lemma, at most 6 of these 27 segments have a length less than 3, i.e., at least 21 segments have a length not less than 3. Without loss of generality, assume the segments $P_{2} P_{3}, P_{2} P_{4}, P_{2} P_{5}, \cdots, P_{2} P_{23}$ have a length not less than 3.

Next, consider the 20 segments $P_{3} P_{4}, P_{3} P_{5}, P_{3} P_{6}, \cdots, P_{2} P_{23}$ starting from $P_{3}$. According to the lemma, at most 6 of these 20 segments have a length less than 3, i.e., at least 14 segments have a length not less than 3. Without loss of generality, assume the segments $P_{3} P_{4}, P_{3} P_{5}, P_{3} P_{6}, \cdots, P_{3} P_{17}$ have a length not less than 3.

Next, consider the 13 segments $P_{4} P_{5}, P_{4} P_{6}, P_{4} P_{7}, \cdots, P_{4} P_{17}$ starting from $P_{4}$. According to the lemma, at most 6 of these 13 segments have a length less than 3, i.e., at least 7 segments have a length not less than 3. Without loss of generality, assume the segments $P_{4} P_{5}, P_{4} P_{6}, P_{4} P_{7}, \cdots, P_{4} P_{11}$ have a length not less than 3.

Thus, we have 5 points $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$. According to the previous discussion, the distance between any two of these 5 points is not less than 3.
Therefore, the conclusion holds.