Does anyone have a solution (e.g. official)

I've found an unofficial solution.

Since the solution uses modular arithmetic, it is convenient to assume that all indices vary from $0$ to $n-1$, not from $1$ to $n$.

To prove that the numbers $a_{ij}$ and $b_i$ satisfy the condition, suppose we take any choice of $x_1,x_2,\dots,x_n \in S$. Let $M$ be the total number of $-1$ among $x_i$. We need to prove that the number of $+1$ among $y_i$ is greater than the number of $-1$ among them if $M$ is even, and is less than it otherwise. For this purpose, we call two indices $i_1$ and $i_2$ coherent if $i_1-i_2 \equiv \pm k \pmod n$. Clearly, there are exactly two indices coherent to a given one, and since $k$ is coprime to $n$, all $n$ indices can be arranged in a circle where every pair of adjacent indices are coherent: \[0,k,2k,3k \bmod n=k-1,\dots,k+1,0.\] Lemma If the indices $i_1$ and $i_2$ are coherent, then at least one of the numbers $y_{i_1}, y_{i_2}$ is equal to $(-1)^M$. Proof The bases $B_{i_1}$ and $B_{i_2}$, corresponding to $i_1$ and $i_2$ respectively, are disjoint sets of $k$ elements each. Let $p$, $q$ and $r$ be the numbers of $-1$ among $x_j$ when $j$ is in $B_{i_1}$, $B_{i_2}$ and is equal to the remaining $(2k+1)$ th index, respectively ($r$ is $0$ or $1$). Then $M=p+q+r$. Since $y_i=\operatorname{sign} \, (n-2M_i)$, where $M_i$ is the number of $-1$ among $a_{ij}x_j$, $y_i=1$ if and only if $2k+1 \ge 2M_i \Leftrightarrow k \ge M_i$. Clearly, $M_{i_1}=(k-p)+q+r$ and $M_{i_2}=p+(k-q)+r$. If $M$ is odd and we assume that $y_{i_1}=y_{i_2}=1$, then it must be $k \ge k-p+q+r$ and $k \ge p+k-q+r$, which implies $p \ge q+r$ and $q \ge p+r$. Adding the last inequalities, we obtain: $0 \ge r$, which impiles $r=0$, $p=q$ and $M=p+q+r=2p$, that contradicts the fact that $M$ is odd. If $M$ is even and we assume that $y_{i_1}=y_{i_2}=-1$, then it must be $k<k-p+q+r$ and $k<p+k-q+r$, which implies $p<q+r$ and $q<p+r$. Adding the last inequalities, we obtain: $0<r$, which impiles $r=1$, $p=q$ and $M=p+q+1=2p+1$, that contradicts the fact that $M$ is even. $\Box$ Now, to finish the proof of the main statement, we first apply the lemma and receive (any) index $i$ for which $y_i=(-1)^M$, then break up the other indices into pairs of coherent ones, and apply the lemma again for each pair. As a result, we obtain more that half of the indices $i$ for which $y_i$ is equal to $(-1)^M$, i.e. to $x_1 x_2 \dots x_n$. $\blacksquare$

A beautiful solution to this hard problem! I was just wondering, how do you think of $a_{ij}$ and $b_i$?

By trial and error method, starting from small values of $n$. But the hardest thing in solving this problem was to find a proof, not $a_{ij}$ and $b_i$.

May I ask, what was the inspiration for the proof?

A sound of artillery shelling outside. But it is was rather a nudge than an inspiration. Seriously, I don't remember any more about that day.

I think this works:

A cool problem! Maybe a little bit hard for problem 1. The answer is yes. Select $b_1 = b_2 = \cdots = b_n = 1$ , and $(a_{ij})$ be \[ a_{ij} = \begin{cases} -1 &, j-i \equiv 0, 1, 2, \dots (n-3)/2 \pmod n \\ 1 &, \text{otherwise} \end{cases} \] We show that it satisfies the condition. Let $y_i' = \sum_{j=1}^n a_{ij} x_j$, $z' = \sum_{i=1}^nb_iy_i$ . For $(x_i) \in S^n$, since $n$ is an odd number, $y_i'$ and $z'$ are odd numbers ,which means none ofthem are zero. So if we replace $(x_i)$ with $(-x_i)$ , all of $y_i$ switch, then $z$ also switch. Hence WLOG, $(x_i)$ has an odd number of $1$, then it suffices to show $z' \ge 0$ 。 Let $s_i$ be the numbers of $1$ among $x_i, x_{i+1}, \dots, x_{i+(n-3)/2}$ , then \[ \begin{aligned} y_i' &= (x_i+x_{i+1}+\cdots +x_{i+(n-1)/2})-(x_{i+(n+1)/2}+\cdots +x_{i-1}) \\ &= x_i+(2s_{i+1} - (n-3)/2)-(2s_{i+(n+1)/2} - (n-3)/2) \\ &= x_i+2(s_{i+1}-s_{i+(n+1)/2}) \end{aligned} \]also \[ \begin{aligned} y_{i+(n+1)/2}' &= (x_{i+(n+1)/2}+\cdots +x_{i+n})-(x_{i+n+1}+\cdots +x_{i+(3n-1)/2}) \\ &= (2s_{i+(n+1)/2} - (n-3)/2)+x_i-(2s_{i+1} - (n-3)/2) \\ &= x_i+2(s_{i+(n+1)/2}-s_{i+1}) \end{aligned} \]Pair up $y_i'$ and $y_{i+(n+1)/2}'$. If both of $y_i', y_{i+(n-1)/2}'$ are negative, then $0 > y_i'+y_{i+(n+1)/2}' = 2x_i$ , hence $x_i = -1$. Since there are an odd number of$1$, we have $s_{i+1}+s_{i+(n+1)/2}$ is odd, so $s_{i+1}-s_{i+(n+1)/2} \neq 0$., and that contradict to $y_i', y_{i+(n+1)/2}' < 0$. Hence at least one of$y_i, y_{i+(n+1)/2}$ is $1$, that is \[ 2\sum_{i=1}^n y_i = \sum_{i=1}^n y_i + y_{i+(n+1)/2} \ge 0\]Hence $z=1$, as desired.