When are $\beta_1, \beta_2, \ldots, \beta_n$ linear independent? Given $n$ linear independent vectors, $\alpha_1, \alpha_2, \ldots, \alpha_n$.
Now, let
$$\beta_1 = \alpha_1 + \alpha_2 + \cdots + \alpha_m$$
$$\beta_2 = \alpha_2 + \alpha_3 + \cdots + \alpha_{m+1}$$
$$\ldots$$
$$\beta_{n-m} = \alpha_{n-m} + \alpha_{n-m+1} + \cdots + \alpha_n$$
$$\beta_{n-m+1} = \alpha_{n-m+1} + \alpha_{n-m+2} + \cdots + \alpha_{n+1}$$
$$\ldots$$
$$\beta_{n-1} = \alpha_{n-1} + \alpha_{n} + \cdots + \alpha_{m-2}$$
$$\beta_{n} = \alpha_{n} + \alpha_{1} + \cdots + \alpha_{m-1}$$
where $1 \lt m \lt n$.
For example, if $n=3$ and $m=2$, then
$$\beta_1 = \alpha_1 + \alpha_2$$
$$\beta_2 = \alpha_2 + \alpha_3$$
$$\beta_3 = \alpha_3 + \alpha_1$$
The question is, to which condition $n$ and $m$ must meet when $\beta_1, \beta_2, \ldots, \beta_n$ are linear independent?
A guess is that $n$ and $m$ must be relatively prime, but I can neither prove or disprove it.

Isn't the implied matrix a circulant matrix?
Then, because of this determinant formula, your problem (for the case of linearly dependent vectors) appears to ask, for given $m < n$

When is it the case that there is a $j \mid n$ such that $\omega_{j}$ is a root of $f_{m} = x^{m-1} + \dots + x + 1$, or equivalently $\phi_{j}$ divides $f_{m}$?

Here $\omega_{j}$ is a primitive $j$-th root of unity, and $\phi_{j}$ is the $j$-th cyclotomic polynomial. 
If $\phi_{j}$ divides $f_{m}$, then $j > 1$ (as $f_{m}(\omega_{1}) = f_{m}(1) = m \ne 0$), and $\phi_{j}$ divides $x^{m} - 1 = (x-1) f_{m}$, so $j \mid m$. Conversely, if $j > 1$ and $j \mid m$, then $\phi_{j}$ divides $x^{j} - 1$ which divides $x^{m} - 1 = (x - 1) f_{m}$, so that $\phi_{j}$ divides $f_{m}$, as $j > 1$.
Therefore the condition for linear dependence appears indeed to be that $\gcd(n, m) \ne 1$.

Addendum.
The key point here is really 
$$
\gcd(x^{n} - 1, x^{k} - 1)
=
x^{\gcd(n, k)} - 1.
$$
