20. Suppose $k \geq 1 / 2 + \sqrt{2 n}$. Consider a point $P$ in $S$. There are at least $k$ points in $S$ having all the same distance to $P$, so there are at least $\binom{k}{2}$ pairs of points $A, B$ with $A P = B P$. Since this is true for every point $P \in S$, there are at least $n \binom{k}{2}$ triples of points $(A, B, P)$ for which $A P = B P$ holds. However,
$$\begin{aligned} n \binom{k}{2} & = n \frac{k(k-1)}{2} \geq \frac{n}{2} \left( \sqrt{2 n} + \frac{1}{2} \right) \left( \sqrt{2 n} - \frac{1}{2} \right) \\ & = \frac{n}{2} \left( 2 n - \frac{1}{4} \right) > n(n-1) = 2 \binom{n}{2} \end{aligned}$$
Since $\binom{n}{2}$ is the number of all possible pairs $(A, B)$ with $A, B \in S$, there must exist a pair of points $A, B$ with more than two points $P_{i}$ such that $A P_{i} = B P_{i}$. These points $P_{i}$ are collinear (they lie on the perpendicular bisector of $A B$), contradicting condition (1).