Let $n$ be a positive integer and $a_1, a_2, \dots , a_{2n}$ mutually distinct integers. Find all integers $x$ satisfying
\[(x - a_1) \cdot (x - a_2) \cdots (x - a_{2n}) = (-1)^n(n!)^2.\]
1. Given the equation:
   \[
   (x - a_1) \cdot (x - a_2) \cdots (x - a_{2n}) = (-1)^n(n!)^2,
   \]
   we need to find all integers \( x \) that satisfy this equation.

2. Notice that the left-hand side is a polynomial of degree \( 2n \) with roots \( a_1, a_2, \ldots, a_{2n} \). The right-hand side is a constant, specifically \( (-1)^n(n!)^2 \).

3. The absolute value of the product of the differences \( |(x - a_1) \cdot (x - a_2) \cdots (x - a_{2n})| \) must equal \( (n!)^2 \). This implies that the differences \( |x - a_i| \) must be such that their product is \( (n!)^2 \).

4. Since \( a_1, a_2, \ldots, a_{2n} \) are distinct integers, the smallest possible values for \( |x - a_i| \) are \( 1, 2, 3, \ldots, n, -1, -2, -3, \ldots, -n \). This is because the product of these values will give us \( (n!)^2 \).

5. Therefore, the set of differences \( (x - a_1), (x - a_2), \ldots, (x - a_{2n}) \) must be \( \{1, 2, 3, \ldots, n, -1, -2, -3, \ldots, -n\} \).

6. To satisfy this, \( x \) must be the average of the \( a_i \)'s. This is because the sum of the differences must be zero:
   \[
   (x - a_1) + (x - a_2) + \cdots + (x - a_{2n}) = 0.
   \]
   Simplifying, we get:
   \[
   2nx - (a_1 + a_2 + \cdots + a_{2n}) = 0,
   \]
   which implies:
   \[
   x = \frac{a_1 + a_2 + \cdots + a_{2n}}{2n}.
   \]

7. Since \( x \) must be an integer, the sum \( a_1 + a_2 + \cdots + a_{2n} \) must be divisible by \( 2n \).

Conclusion:
The integer \( x \) that satisfies the given equation is:
\[
x = \frac{a_1 + a_2 + \cdots + a_{2n}}{2n}.
\]

The final answer is \( \boxed{ x = \frac{a_1 + a_2 + \cdots + a_{2n}}{2n} } \).