Find all $n$-tuples of real numbers $(x_1, x_2, \dots, x_n)$ such that, for every index $k$ with $1\leq k\leq n$, the following holds:

\[ x_k^2=\sum\limits_{\substack{i < j \\ i, j\neq k}} x_ix_j \]

[i]Proposed by Oriol Solé[/i]
1. **Identify the given condition:**
   For every index \( k \) with \( 1 \leq k \leq n \), we have:
   \[
   x_k^2 = \sum_{\substack{i < j \\ i, j \neq k}} x_i x_j
   \]

2. **Consider the case where all elements are equal:**
   Suppose \( x_1 = x_2 = \cdots = x_n = x \). Then the given condition becomes:
   \[
   x^2 = \sum_{\substack{i < j \\ i, j \neq k}} x^2
   \]
   The number of pairs \((i, j)\) with \( i < j \) and \( i, j \neq k \) is \(\binom{n-1}{2}\). Therefore, the equation simplifies to:
   \[
   x^2 = \binom{n-1}{2} x^2
   \]
   This implies:
   \[
   x^2 \left( \binom{n-1}{2} - 1 \right) = 0
   \]
   Hence, either \( x = 0 \) or \(\binom{n-1}{2} = 1\). The latter condition simplifies to:
   \[
   \frac{(n-1)(n-2)}{2} = 1 \implies (n-1)(n-2) = 2
   \]
   Solving this quadratic equation, we get \( n = 3 \).

3. **Consider the case where not all elements are equal:**
   Suppose \( x_k \neq x_\ell \) for some indices \( k \) and \( \ell \). Then:
   \[
   x_k^2 = \sum_{\substack{i < j \\ i, j \neq k}} x_i x_j
   \]
   and
   \[
   x_\ell^2 = \sum_{\substack{i < j \\ i, j \neq \ell}} x_i x_j
   \]
   Subtracting these two equations, we get:
   \[
   x_k^2 - x_\ell^2 = \sum_{\substack{i < j \\ i, j \neq k}} x_i x_j - \sum_{\substack{i < j \\ i, j \neq \ell}} x_i x_j
   \]
   This can be rewritten as:
   \[
   (x_k - x_\ell)(x_k + x_\ell) = (x_\ell - x_k) \left( \sum_{i \not\in \{k, \ell\}} x_i \right)
   \]
   Since \( x_k \neq x_\ell \), we can divide by \( x_k - x_\ell \) to get:
   \[
   -x_k - x_\ell = \sum_{i \not\in \{k, \ell\}} x_i
   \]
   Summing over all \( k \) and \( \ell \), we get:
   \[
   \sum_{i=1}^n x_i = 0
   \]

4. **Compute the sum of squares:**
   \[
   \sum_{m=1}^n x_m^2 = \sum_{m=1}^n \sum_{\substack{i < j \\ i, j \neq m}} x_i x_j
   \]
   Each pair \((i, j)\) appears in the sum for exactly \( n-2 \) values of \( m \). Therefore:
   \[
   \sum_{m=1}^n x_m^2 = (n-2) \sum_{i < j} x_i x_j
   \]
   Also, since \( \sum_{i=1}^n x_i = 0 \):
   \[
   0 = \left( \sum_{m=1}^n x_m \right)^2 = \sum_{m=1}^n x_m^2 + 2 \sum_{i < j} x_i x_j
   \]
   Thus:
   \[
   \sum_{i < j} x_i x_j = -\frac{1}{2} \sum_{m=1}^n x_m^2
   \]
   Substituting this back, we get:
   \[
   \sum_{m=1}^n x_m^2 = (n-2) \left( -\frac{1}{2} \sum_{m=1}^n x_m^2 \right)
   \]
   Simplifying, we find:
   \[
   \frac{n}{2} \sum_{m=1}^n x_m^2 = 0
   \]
   Hence:
   \[
   \sum_{m=1}^n x_m^2 = 0
   \]
   This implies \( x_i = 0 \) for all \( i \), contradicting \( x_k \neq x_\ell \).

Thus, the only solutions are \( (0, 0, \dots, 0) \) for any \( n \) and \( (x, x, x) \) for \( n = 3 \).

The final answer is \( \boxed{ (0, 0, \dots, 0) } \) for any \( n \) and \((x, x, x)\) for \( n = 3 \).