$p$ is an odd prime number. Find all $\frac{p-1}2$-tuples $\left(x_1,x_2,\dots,x_{\frac{p-1}2}\right)\in \mathbb{Z}_p^{\frac{p-1}2}$ such that
$$\sum_{i = 1}^{\frac{p-1}{2}} x_{i} \equiv \sum_{i = 1}^{\frac{p-1}{2}} x_{i}^{2} \equiv \cdots \equiv \sum_{i = 1}^{\frac{p-1}{2}} x_{i}^{\frac{p - 1}{2}} \pmod p.$$

[i]Proposed by Ali Partofard[/i]
1. **Restate the problem and initial conditions:**
   Let \( p \) be an odd prime number. We need to find all \(\frac{p-1}{2}\)-tuples \((x_1, x_2, \dots, x_{\frac{p-1}{2}}) \in \mathbb{Z}_p^{\frac{p-1}{2}}\) such that:
   \[
   \sum_{i=1}^{\frac{p-1}{2}} x_i \equiv \sum_{i=1}^{\frac{p-1}{2}} x_i^2 \equiv \cdots \equiv \sum_{i=1}^{\frac{p-1}{2}} x_i^{\frac{p-1}{2}} \pmod{p}.
   \]

2. **Simplify the problem:**
   Write \( p = 2k + 1 \). If \( k = 1 \), then \( p = 3 \), and any singleton will satisfy the condition. Suppose \( p > 3 \), i.e., \( k > 1 \).

3. **Identify potential solutions:**
   Consider tuples \((x_1, x_2, \cdots, x_k) \in \{0, 1\}^k\). These tuples satisfy the given conditions. We need to show that these are the only solutions.

4. **Polynomial approach:**
   For any polynomial \( P \) of degree \(\leq \frac{p-1}{2}\) with the sum of coefficients congruent to 0 modulo \( p \), we have:
   \[
   \sum P(x_i) \equiv P(0) \cdot (k - t) \pmod{p},
   \]
   where \( t = \sum x_i \).

5. **Construct a specific polynomial:**
   Consider the polynomial \( P(x) = (b - x)^k - (b-1)^k \). This polynomial is divisible by \( x-1 \), so the sum of its coefficients is 0. Therefore:
   \[
   \left(\sum (b - x_i)^k \right) - k (b-1)^k \equiv \left(b^k - (b-1)^k \right) \cdot (k - t) \pmod{p},
   \]
   for all integers \( b \).

6. **Use properties of Legendre symbols:**
   Notice that \( a^k \equiv \left( \frac{a}{p} \right) \) for all integers \( a \). Thus, whenever \( \left( \frac{b}{p} \right) = \left( \frac{b-1}{p} \right) \), we have:
   \[
   \left(\sum \left(\frac{b-x_i}{p} \right) \right) \equiv k \cdot \left( \frac{b-1}{p} \right) \pmod{p}.
   \]

7. **Bounding and implications:**
   Since \( \left( \frac{b-x_i}{p} \right) \in \{-1, 0, 1\} \), simple bounding yields:
   \[
   \left( \frac{b - x_i}{p} \right) = \left( \frac{b-1}{p} \right),
   \]
   for all \( 1 \le i \le k \).

8. **Analyze quadratic residues:**
   For every \( 1 \le i \le k \) and integer \( a \), if \( \left( \frac{a(a-1)}{p} \right) = 1 \), then:
   \[
   \left( \frac{a(a-x_i)}{p} \right) = 1.
   \]
   We claim that this implies \( x_i \in \{0, 1\} \).

9. **Contradiction argument:**
   Suppose \( x_i \notin \{0, 1\} \). It is known that for a quadratic \( f(x) = \alpha x^2 + \beta x + \gamma \) not a perfect square in \( \mathbb{Z}_p[x] \):
   \[
   \sum_{i=0}^{p-1} \left( \frac{f(i)}{p} \right) = -1.
   \]
   For \( f_1(x) = x^2 - x \) and \( f_2(x) = x^2 - kx \), there are precisely two roots for each, \(\frac{p-3}{2}\) residues for which they evaluate to a nonzero square modulo \( p \), and \(\frac{p-1}{2}\) for which they evaluate to a non-quadratic residue.

10. **Summing over residue classes:**
    From the above, for all \( a \notin \{0, 1, k\} \):
    \[
    \left( \frac{a(a-1)}{p} \right) = \left( \frac{a(a-k)}{p} \right) \Rightarrow \left( \frac{(a-1)(a-k)}{p} \right) = 1.
    \]
    Summing over all residue classes and applying the result gives a contradiction.

11. **Conclusion:**
    Hence, our assumption that \( x_i \notin \{0, 1\} \) was incorrect. Therefore, the only solutions are \((x_1, x_2, \cdots, x_k) \in \{0, 1\}^k\).

\(\blacksquare\)

The final answer is \( \boxed{ (x_1, x_2, \cdots, x_k) \in \{0, 1\}^k } \).