The $n$-tuples are $(0, 0, \dots, 0)$ for each $n$, and the triplet $(x, x, x)$ for each real $x$, which are clearly solutions. We claim that all the elements of the tuple must be equal. Suppose by contradiction that $x_k \neq x_{\ell}$ for some indices $k$ and $\ell$. Let us observe then that $$x_k^2 = \sum\limits_{\substack{i < j \\ i, j \neq k}} x_ix_j = \sum\limits_{\substack{i < j \\ i, j \not \in \{k, \ell\}}} x_ix_j + \sum\limits_{i \not \in \{k, \ell\}} x_ix_\ell = \sum\limits_{\substack{i < j \\ i, j \not \in \{k, \ell\}}} x_ix_j + x_{\ell}\left(\sum\limits_{i \not \in \{k, \ell\}} x_i\right)$$ In a similar way we obtain $$x_{\ell}^2 = \sum\limits_{\substack{i < j \\ i, j \not \in \{k, \ell\}}} x_ix_j + x_{\ell}\left(\sum\limits_{i \not \in \{k, \ell\}} x_i\right)$$ Substracting both equations we have $$(x_k - x_\ell)(x_k + x_\ell) = x_k^2 - x_{\ell}^2 = (x_{\ell} - x_k)\left(\sum\limits_{i \not \in \{k, \ell\}} x_i\right)$$ And since $x_k - x_{\ell} \neq 0$ we infer dividing by $x_k - x_{\ell}$ that $$- x_k - x_{\ell} = \sum\limits_{i \not\in \{k, \ell\}} x_i \implies \sum_{i = 1}^n x_i = 0$$ Let us compute the value of the sum $$\sum_{m = 1}^n x_m^2 = \sum_{m = 1}^n \sum\limits_{\substack{i < j \\ i, j \neq m}} x_ix_j$$ We can see that for each pair $(i, j)$ with $i < j$ there exist exactly $(n - 2)$ values of $m$ such that $x_ix_j$ appears in the expression of $x_m^2$. We then infer that $$\sum_{m = 1}^n x_m^2 = (n - 2)\left( \sum_{i < j} x_ix_j \right)$$ On the other hand, we can see that $$0 = \left(\sum_{m = 1}^n x_m\right)^2 = \sum_{m = 1}^n x_m^2 + 2\sum_{i < j} x_ix_j$$ So $$\sum_{i < j} x_ix_j = -\frac{1}{2} \sum_{m = 1}^n x_m^2$$ Substituting in the expression obtained before, we obtain $$\frac{n}{2} \sum_{m = 1}^n x_m^2 = 0$$ And the sum of the squares of the $x_i$ is equal to $0$, which implies that $x_i = 0$ for all $i$, contradicting that $x_k \neq x_{\ell}$. Thus, it follows that $x_1 = x_2 = \dots = x_n = x$, so the expression for the squares of the $x_i$ becomes $$x^2 = \sum\limits_{\substack{i < j \\ i, j \neq k}} x^2$$ We can easily see that the number of pairs $(i, j)$ with $i < j$ and $i, j \neq k$ is equal to $\binom{n - 1}{2}$. This corresponds to choose twp different indices from the set $\{1, 2, \dots, k - 1, k + 1, \dots, n - 1, n\}$. The last equation becomes $$\binom{n - 1}{2}x^2 = x^2 \implies \left(\binom{n - 1}{2} - 1\right)x^2 = 0$$ from which we finally get $x = 0$ or $n = 3$.