IMC 2018 · Day 2 · P6

First we prove that if $2n + 1 \le k$ then no sequence $v_1, \dots, v_k$ of vectors can satisfy the condition. Suppose to the contrary that $v_1, \dots, v_k$ are vectors with the required property and consider the vectors $$v_1, v_3, v_5, \dots, v_{2n+1}.$$ By the condition these $n + 1$ vectors should be pairwise orthogonal, but this is not possible in $\mathbb{R}^n$.

Next we show a possible construction for every pair $k, n$ of positive integers with $2n \ge k$. Take an orthogonal basis $(e_1, \dots, e_n)$ of $\mathbb{R}^n$ and consider the vectors $$v_1 = v_2 = e_1, \quad v_3 = v_4 = e_2, \quad \dots, \quad v_{2n-1} = v_{2n} = e_n.$$ For every pair $(i, j)$ of indices with $1 \le i, j \le 2n$ and $|i - j| > 1$ the vectors $v_i$ and $v_j$ are distinct basis vectors, so they are orthogonal. Evidently the subsequence $v_1, v_2, \dots, v_k$ also satisfies the same property.

Hence, such a sequence of vectors exists if and only if $2n \ge k$; that is, for a fixed $k$, the smallest suitable $n$ is $\left\lceil \dfrac{k}{2} \right\rceil$.