It seems that the shortlist had only one actual algebra problem (the one which came out at the competition ), in my opinion this one has more of a number theory flavour because of the use of Bezout or similar ideas. If $k$ is not relatively prime with $n$, say with gcd$(k,n) = d > 1$, then gluing blocks of length $d$ of the form $1, 0, 0, \ldots, 0, (-1)$ (we allow the number of $0$s to be zero, works for $d=2$) gives a counterexample. If $k$ is relatively prime with $n$, then by Bezout there is $x$ with $kx \equiv 1 \pmod n$, so since $x_i = x_{i+k}$ for all $i$ we get $x_i = x_{i+kx}$ for all $i$ and hence $x_i = x_{i+1}$, impliying that all $x$-s are equal - to zero, because the sum of each $k$ consecutive is $0$.

Consider all indices modulo $n$, except when noted otherwise. We claim that the only $k$ that work are those that satisfy $\gcd(n,k) = 1$. Suppose that $\gcd(n,k) = d>1$. Then, we claim that, if we have $x_i+x_{i+1}+\cdots + x_{i+d-1}=0$, then we have $x_i + x_{i+1} + \cdots + x_{i+k-1} = 0$. Notice that since $d\mid k$, the integers $i, i+1, \dots, i+k-1$, when considered modulo $d$, are several permutations of $0,1,\dots, d-1$. But since $x_i+x_{i+1}+\cdots + x_{i+d-1}=0$, then clearly we must have the desired. From here, it is not hard to see that we need not have $x_1=x_2=\cdots x_n = 0$. For example, since $k>1$, we could have: \[x_i = \begin{cases} -1&\text{if } d\mid i \\ \tfrac{1}{d-1} &\text{if } d\nmid i.\end{cases}\]Now, we show that if $\gcd(n,k)=1$, then $x_1=x_2=\cdots = x_n$. Note that by \begin{align*} x_i+x_{i+1}+\cdots + x_{i+k-1} &= 0 \\ x_{i+1} + x_{i+2} + \cdots + x_{i+k} &= 0,\end{align*}we have that $x_i = x_{i+k}$, or $x_i = x_{i+rk}$, for any integer $r$. But since $k$ is relatively prime to $n$, it is clear that $i+rk$ can take on any value modulo $n$, which implies that $x_1=x_2=\cdots = x_n$. It follows that $x_1=x_2=\cdots = x_n = 0$, as desired. $\blacksquare$