11. (CAN 1) Let $n$ be a natural number and $a_{1}, a_{2}, \ldots, a_{2 n}$ mutually distinct integers. Find all integers $x$ satisfying  
$$ \left(x-a_{1}\right) \cdot\left(x-a_{2}\right) \cdots\left(x-a_{2 n}\right)=(-1)^{n}(n!)^{2} $$
11. Suppose that an integer \( x \) satisfies the equation. Then the numbers \( x - a_1, x - a_2, \ldots, x - a_{2n} \) are \( 2n \) distinct integers whose product is \( 1 \cdot (-1) \cdot 2 \cdot (-2) \cdots n \cdot (-n) \). From here it is obvious that the numbers \( x - a_1, x - a_2, \ldots, x - a_{2n} \) are some reordering of the numbers \( -n, -n+1, \ldots, -1, 1, \ldots, n-1, n \). It follows that their sum is 0, and therefore \( x = \left( a_1 + a_2 + \cdots + a_{2n} \right) / 2n \). This is the only solution if \( \{ a_1, a_2, \ldots, a_{2n} \} = \{ x - n, \ldots, x - 1, x + 1, \ldots, x + n \} \) for some \( x \in \mathbb{N} \). Otherwise there is no solution.