Let $\mathbb R$ be the real numbers. Set $S = \{1, -1\}$ and define a function $\operatorname{sign} : \mathbb R \to S$ by
\[ \operatorname{sign} (x) = 
		\begin{cases}
			1 & \text{if } x \ge 0; \\
			-1 & \text{if } x < 0.
		\end{cases} 
\] Fix an odd integer $n$.  Determine whether one can find $n^2+n$ real numbers $a_{ij}, b_i \in S$ (here $1 \le i, j \le n$) with the following property: Suppose we take any choice of $x_1, x_2, \dots, x_n \in S$ and consider the values \begin{align*}
		y_i &= \operatorname{sign} \left( \sum_{j=1}^n a_{ij} x_j \right), \quad \forall 1 \le i \le n; \\
		z &= \operatorname{sign} \left( \sum_{i=1}^n y_i b_i \right)
	\end{align*} Then $z=x_1 x_2 \dots x_n$.
1. **Restate the Problem:**
   We need to determine if there exist \( n^2 + n \) real numbers \( a_{ij}, b_i \in S \) (where \( S = \{1, -1\} \) and \( 1 \le i, j \le n \)) such that for any choice of \( x_1, x_2, \dots, x_n \in S \), the following holds:
   \[
   y_i = \operatorname{sign} \left( \sum_{j=1}^n a_{ij} x_j \right), \quad \forall 1 \le i \le n
   \]
   \[
   z = \operatorname{sign} \left( \sum_{i=1}^n y_i b_i \right)
   \]
   and \( z = x_1 x_2 \dots x_n \).

2. **Assume \( n = 2k + 1 \) (since \( n \) is an odd integer):**
   We will construct the numbers \( a_{ij} \) and \( b_i \) to satisfy the given conditions.

3. **Define \( b_i = 1 \) for all \( i \):**
   This simplifies the problem as we only need to focus on the \( a_{ij} \) values.

4. **Construct \( a_{ij} \) using modular arithmetic:**
   - Calculate \( (j - i) \mod n \).
   - If the result is less than \( k \), set \( a_{ij} = -1 \).
   - Otherwise, set \( a_{ij} = 1 \).

5. **Verify the construction:**
   - For any \( i \), there are exactly \( k \) values of \( j \) such that \( a_{ij} = -1 \). These \( j \) values form the initial interval \(\{0, 1, 2, \dots, k-1\}\) shifted (and rotated if needed) to the right by \( i \).

6. **Lemma:**
   If indices \( i_1 \) and \( i_2 \) are coherent (i.e., \( i_1 - i_2 \equiv \pm k \pmod n \)), then at least one of \( y_{i_1} \) or \( y_{i_2} \) is equal to \( (-1)^M \), where \( M \) is the number of \( -1 \) among \( x_i \).

7. **Proof of Lemma:**
   - Let \( p \), \( q \), and \( r \) be the numbers of \( -1 \) among \( x_j \) when \( j \) is in \( B_{i_1} \), \( B_{i_2} \), and the remaining \((2k+1)\)th index, respectively.
   - \( M = p + q + r \).
   - \( y_i = \operatorname{sign} (n - 2M_i) \), where \( M_i \) is the number of \( -1 \) among \( a_{ij} x_j \).
   - \( y_i = 1 \) if and only if \( 2k + 1 \ge 2M_i \Leftrightarrow k \ge M_i \).
   - \( M_{i_1} = (k - p) + q + r \) and \( M_{i_2} = p + (k - q) + r \).

8. **Case Analysis:**
   - If \( M \) is odd and \( y_{i_1} = y_{i_2} = 1 \), then \( k \ge k - p + q + r \) and \( k \ge p + k - q + r \), implying \( p \ge q + r \) and \( q \ge p + r \). Adding these gives \( 0 \ge r \), which implies \( r = 0 \), \( p = q \), and \( M = p + q + r = 2p \), contradicting \( M \) being odd.
   - If \( M \) is even and \( y_{i_1} = y_{i_2} = -1 \), then \( k < k - p + q + r \) and \( k < p + k - q + r \), implying \( p < q + r \) and \( q < p + r \). Adding these gives \( 0 < r \), which implies \( r = 1 \), \( p = q \), and \( M = p + q + 1 = 2p + 1 \), contradicting \( M \) being even.

9. **Conclusion:**
   - By applying the lemma, we can find an index \( i \) for which \( y_i = (-1)^M \).
   - Break up the other indices into pairs of coherent ones and apply the lemma again for each pair.
   - This results in more than half of the indices \( i \) for which \( y_i \) is equal to \( (-1)^M \), i.e., \( x_1 x_2 \dots x_n \).

The final answer is \( \boxed{ z = x_1 x_2 \dots x_n } \).