If $v_1,v_2,\dots,v_n$ are linearly independent then what about $v_1+v_2,v_2+v_3,\dots,v_{n-1}+v_n,v_n+v_1$? If $v_{1},v_{2},...v_{n}$ are linearly independent then are the following too?
$v_{1}+v_{2},v_{2}+v_{3},...v_{n-1}+v_{n},v_{n}+v_{1}$
I tried summing them to give 0 but had no success.

This is not in general linearly independent. If $n$ is odd, these are linearly independent, but if $n$ is even, they are not in general. For example: $(1,0,0,0),(0,1,0,0), (0,0,1,0)$ and $(0,0,0,1)$, we have $(1,1,0,0)+(0,0,1,1)-(0,1,1,0)-(1,0,0,1) = (0,0,0,0)$.
If $n$ is odd, then if $v_1, v_2, \ldots , v_n$ is linearly independent and $c_1(v_1+v_2) + \ldots + c_n(v_n + v_1) = 0$, then $$ (c_1+c_n)v_1 + (c_1+c_2)v_2 + \ldots + (c_{n-1} + c_n)v_n = 0.$$ It is easy to see that $c_i = -c_{i+1}$ and $c_i = c_{i+2}$ for all $i$ (for all $i = 1,2,\ldots, n-2$). Now, if $n=2k+1$, then we have $c_1 = c_3 = \ldots = c_{2k+1}$ and clearly $c_{2k+1} = c_2$, thus $c_1 = c_2= c_3 = \ldots = c_{2k+1}$ and since $c_i = -c_{i+1}$, this shows that $c_i= 0$ for each $i \in \{1,2,\ldots, n\}$.
