An integer $n \geq 3$ is given. We call an $n$-tuple of real numbers $(x_1, x_2, \dots, x_n)$ [i]Shiny[/i] if for each permutation $y_1, y_2, \dots, y_n$ of these numbers, we have
$$\sum \limits_{i=1}^{n-1} y_i y_{i+1} = y_1y_2 + y_2y_3 + y_3y_4 + \cdots + y_{n-1}y_n \geq -1.$$
Find the largest constant $K = K(n)$ such that
$$\sum \limits_{1 \leq i < j \leq n} x_i x_j \geq K$$
holds for every Shiny $n$-tuple $(x_1, x_2, \dots, x_n)$.
We need to find the largest constant \( K = K(n) \) such that for every Shiny \( n \)-tuple \((x_1, x_2, \dots, x_n)\), the inequality
\[ \sum_{1 \leq i < j \leq n} x_i x_j \geq K \]
holds. We claim that the answer is \( \boxed{K(n) = \frac{1-n}{2}} \) for all \( n \geq 3 \).

1. **Constructing a Shiny \( n \)-tuple:**
   Consider the \( n \)-tuple \((x_1, x_2, \dots, x_n) = (a, a, \dots, a, -\frac{1}{2a})\). We need to check if this \( n \)-tuple is Shiny. For any permutation \((y_1, y_2, \dots, y_n)\) of these numbers, we have:
   \[ \sum_{i=1}^{n-1} y_i y_{i+1} \geq -1. \]
   Since \( y_n = -\frac{1}{2a} \) and the rest are \( a \), the sum becomes:
   \[ (n-2)a^2 + a \left(-\frac{1}{2a}\right) + a \left(-\frac{1}{2a}\right) = (n-2)a^2 - \frac{1}{2} - \frac{1}{2} = (n-2)a^2 - 1. \]
   For this to be \(\geq -1\), it is clear that the \( n \)-tuple is Shiny.

2. **Finding the lower bound for \( K \):**
   The sum of all pairwise products for this \( n \)-tuple is:
   \[ \sum_{1 \leq i < j \leq n} x_i x_j = \binom{n-1}{2}a^2 + \sum_{i=1}^{n-1} a \left(-\frac{1}{2a}\right) = \binom{n-1}{2}a^2 + (n-1)\left(-\frac{1}{2}\right). \]
   Simplifying, we get:
   \[ \binom{n-1}{2}a^2 - \frac{n-1}{2}. \]
   As \( a \to 0 \), the term \(\binom{n-1}{2}a^2 \to 0\), and we are left with:
   \[ -\frac{n-1}{2}. \]
   This shows that \( K \leq \frac{1-n}{2} \).

3. **Proving the inequality holds for \( K = \frac{1-n}{2} \):**
   We need to show that for any Shiny \( n \)-tuple \((x_1, x_2, \dots, x_n)\), the inequality:
   \[ \sum_{1 \leq i < j \leq n} x_i x_j \geq \frac{1-n}{2} \]
   holds. We will consider two cases:

   **Case 1: There are exactly \(\frac{n}{2}\) positive numbers and \(\frac{n}{2}\) negative numbers.**
   Without loss of generality, let \( x_1, x_3, \dots, x_{n-1} < 0 \) and \( x_2, x_4, \dots, x_n > 0 \). We extend indices by \( x_i = x_{i-n} \) for all \( i > n \). Adding all cyclic variants of the assertion gives:
   \[ y_1 y_2 + y_2 y_3 + \cdots + y_{n-1} y_n + y_n y_1 \geq -\frac{n}{n-1}. \]
   Plugging in \((y_1, y_2, \dots, y_n) = (x_1, x_{2k+2}, x_3, x_{2k+4}, \dots, x_{n-1}, x_{2k+n})\), we get:
   \[ x_1 (x_{2k} + x_{2k+2}) + x_3 (x_{2k+2} + x_{2k+4}) + \cdots + x_{n-1} (x_{2k+n-2} + x_{2k+n}) \geq -\frac{n}{n-1}. \]
   Summing up from \( k = 0, 2, 4, \dots, n \), we get:
   \[ \sum_{1 \leq i, j \leq n, 2 \mid i, 2 \nmid j} x_i x_j \geq -\frac{n^2}{4(n-1)}. \]
   Hence,
   \[ \sum_{1 \leq i < j \leq n} x_i x_j \geq \sum_{1 \leq i, j \leq n, 2 \mid i, 2 \nmid j} x_i x_j \geq -\frac{n^2}{4(n-1)} \geq \frac{1-n}{2}. \]

   **Case 2: The number of positive and negative reals are different.**
   In this case, at least one of \( y_1 y_2, y_2 y_3, \dots, y_{n-1} y_n, y_n y_1 \) must be non-negative. Hence,
   \[ y_1 y_2 + y_2 y_3 + \cdots + y_{n-1} y_n + y_n y_1 \geq -1. \]
   Summing all inequalities of the previous form gives the desired result.

\(\blacksquare\)