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.

Suppose there are $\lambda_1,\dotsc,\lambda_n\in \mathbb{R}$ with
$$\lambda_1(v_1+v_2)+\dotsc+\lambda_n(v_n+v_1)=0$$
If $n$ is even we fix $\lambda\in K\setminus\{0\}$ and choose $\lambda_i=(-1)^{i}\lambda$ so for $n$ even the vectors aren't linear independent.
For $n$ odd we get from the above equation
$$(\lambda_1+\lambda_n)v_1+\dotsc +(\lambda_{n-1}+\lambda_n)v_n=0$$
and by linear independence of $v_1,\dotsc,v_n$ we get
$$\lambda_1=-\lambda_n, \lambda_2=-\lambda_1,\dotsc, \lambda_n=-\lambda_{n-1}$$
So we get $-\lambda_1=\lambda_n=(-1)^{n-1}\lambda_1=\lambda_1$ and so $\lambda_1=0$. It follows $\lambda_i=0$ for all $i=1,\dotsc,n$ and so the vectors are linear independent in this case.
