Two posts in french have already appeared on this topic, showing that the maximal number of unit vectors having pairwise negative (resp. non-positive) inner products grow linearly wit the dimension of the space $\mathbb{R}^n$ they live in. This is good fun; however, as mentioned in a comment, it becomes even better fun when one bothers asking, for $\epsilon >0$ how big a family of unit vectors $v_1,\cdots,v_k\in \mathbb{R}^n$ can be if…

1. … for all $i\neq j$, one requires $v_i\cdot v_j<\epsilon$;
2. … for all $i\neq j$, one requires $v_i\cdot v_j<-\epsilon$;

Let’s deal with the first question. This involves the remarkable fact that on the sphere of very large dimension, almost all the area is concentrated on a small band near the equator (or any equator). This is the phenomenon called concentration of measure ; so pick a vector $v_1\in\mathbb{S}^n$. To continue your family, you have to pick a vector outside the spherical ball $B=\{x\mid x\cdot v_1\geq\epsilon\}$. So the concentration of measure implies that the forbidden area is of order $C\exp(-cn\epsilon^2)$. When you have picked already $k$ vectors, the forbidden area for the choice of the next one is at most $kC\exp(-cn\epsilon^2)$ which is strictly less than 1 until $k=C^{-1}\exp(cn\epsilon^2)$. Then not all the sphere is forbidden, so you can accommodate certainly $C^{-1}\exp(cn\epsilon^2)$: exponentially (in the dimension) many instead of linearly.

The second one is easier. Indeed,

$0\leq \lVert \sum_{i=1}^k v_i \rVert^2=k+2\sum_{i

which implies $k\leq \epsilon^{-1}+1$ so we get a constant bound, certainly much lower than $n+1$ for large dimensions.