Let $d$ and $n$ be integers. For $i \in \lbrace 1,\dots,n \rbrace$ let $x_i \in \mathbb{R}^d$ be a vector such that $\lVert x \rVert=1 $. For a fixed $1/2 < \alpha \leq 1$, assume we have $\lVert \sum_{i=1}^{n}x_i \rVert \geq \alpha n$ where $\lVert \cdot \rVert$ denotes $\ell_2$ norm.
Question:
Let $w = \frac{ \sum_{i=1}^{n}x_i}{\lVert \sum_{i=1}^{n}x_i \rVert}$. Define the spherical cap of degree $\theta \in [0,\pi/2]$ as follows:
$$ C_{\theta} = \lbrace x \in \mathbb{R}^d: \lVert x \rVert = 1 ~ \text{and} ~ \langle x , w \rangle \geq \cos(\theta) \rbrace. $$
My goal is to find maximum $\cos(\theta)$ such that at least half, i.e., $> n/2 $, of $x_i$s belong to $C_{\theta}$.
Corner Case:
if $\alpha=1$ then it is easy to see that all the vectors are in the direction of $w$. Therefore, setting $\theta=0$ we can see that the number of vectors that belong to $C_{\theta}$ is $n$.
Not an answer, just a remark about the other extreme, of $\alpha = \frac{1}{2}$.
Note that if $n = 2m$ then we can arrange $m$ of the vectors on the equator, spacing them out to look like the $m$'th roots of unity, so that when we add all the vectors on the equator, we get the $0$ vector. The other half can be arranged to all be aligned with one another and orthogonal to the first $m$, and thus the sum $$ \left\lVert \sum_{i=1}^{n}x_i \right\rVert = m = \frac{n}{2} $$
Then if we reasonably interpret "A large fraction" as more than half of the $x_i$'s, then the minimum $\cos(\theta)$ is exactly $0$, since we need to include all of the vectors on the equator, which are all orthogonal to $w$.