At first, if $\beta=\pi/2$, we know that at most $n$ such vectors exist, that is, orthogonal vectors. It's obvious that the number of vectors is influenced by the angle $\beta$. Assume we've already found $k$ vectors satisfying our condition, denoted as $e_1,e_2,\dots,e_k$ normalized. The inner product $(e_1+e_2+\dots e_k, e_1+e_2+\dots+e_k) = k+(k^2-k)*\cos(\beta)\geq 0$, that is, $1+(k-1)\cos(\beta)\geq 0$. if $\pi/2 < \beta < \pi$, then $k \leq 1-1/\cos(\beta)$.
But I can't figure out exactly how many such vectors exist. So the best answer may be,for example, if we say that there exist $n$ such vectors, then could we just propose a construction method finding $n$ such vectors out? I have thought about an "improved" Gram–Schmidt process. While the Gram–Schmidt process orthogonally projects vectors, can we "improve" it to obliquely projects vectors according to the angle $\beta$? I'am not sure whether it works and I don't see people having done this.
I think this is an interesting question. So any body here get good ideas? Welcome to discuss with me! Thank you for your participation.
For many $\beta$, one can always find $n$ vectors by starting from the standard basis and shearing it along the main diagonal, i.e. you use $v_k=e_k+t\cdot d$ for suitable $t$ where $d=(1,1,\ldots, 1)$. Then $\langle v_i,v_i\rangle=1+2t+nt^2$ and $\langle v_i,v_j\rangle=2t+nt^2$ for $i\ne j$, so $\cos(v_i,v_j)=\frac{2t+nt^2}{1+2t+nt^2}=1-\frac 1{1+2t+nt^2}$. This covers the range $1>\cos\beta \ge -\frac{1}{n-1}$. In general, it is not possible to add an $(n+1)$st vector, but for $\cos \beta=-\frac1n$ a negative multiple of $d$ does the trick. Of course with $\cos\beta=1$ we can have arbitrarily many such vectors. For $\cos\beta <-\frac1{n-1}$ we are out of luck, i.e. we can only find fewer than $n$ vectors by finding $m<n$ with $\cos \beta\ge -\frac 1{m-1}$ and use $m$ vectors from an $m$-dimensional subspace.
EDIT: I just notice that I only constructed a set of $n$ vectors, but did not really show that more is not possible. Assume $w_1,\ldots , w_m$ are unit vectors in $n$-space with $\langle w_i,w_j\rangle =\cos \beta$ for $i\ne j$. Let $w=\sum w_i$. Then $0\le\langle w,w\rangle=m\cdot 1+m(m-1)\cdot \cos \beta$ shows that $\cos \beta\ge-\frac1{m-1}$. If $\cos \beta>-\frac1{m-1}$, the $w_i$ are linearly independant (so that we must have $m\le n$). Indeed, the determinant of the matrix with $1$ on the diagonal and $c$ everywhere else is nonzero if $c\ne\frac1{n-1}$ because we can combine the rows into a nonzero multiple of the all-ones vector $d$ and use this to turn each row into a row of the identity matrix.