$n>1$ is an integer. The first $n$ primes are $p_1=2,p_2=3,\dotsc, p_n$. Set $A=p_1^{p_1}p_2^{p_2}...p_n^{p_n}$.
Find all positive integers $x$, such that $\dfrac Ax$ is even, and $\dfrac Ax$ has exactly $x$  divisors
Given the problem, we need to find all positive integers \( x \) such that \(\frac{A}{x}\) is even and \(\frac{A}{x}\) has exactly \( x \) divisors, where \( A = p_1^{p_1} p_2^{p_2} \cdots p_n^{p_n} \) and \( p_i \) are the first \( n \) primes.

1. **Form of \( x \)**:
   Since \( x \) must be a divisor of \( A \), it can be written as \( x = \prod_{i=1}^n p_i^{a_i} \) where \( 0 \le a_i \le p_i \).

2. **Condition for \(\frac{A}{x}\) to be even**:
   We have \( v_2(A) = p_1 = 2 \). For \(\frac{A}{x}\) to be even, \( v_2(x) \) must be either 0 or 1.

3. **Case I: \( v_2(x) = 0 \)**:
   - This implies \( d(A/x) \) is odd, so \( A/x \) must be a perfect square.
   - Therefore, \( a_i \) must be odd for all \( i \ge 2 \).
   - Since \( v_2(A/x) = 2 \), we have \( 2 + 1 = 3 \mid d(A/x) = x \).
   - This implies \( v_3(A/x) = 2 \) or \( 0 \).
   - If \( v_3(A/x) = 2 \), then \( 3 \mid p_2 - a_2 + 1 \) and \( 3 \mid p_1 - a_1 + 1 \), leading to \( 3^2 \mid d(A/x) = x \), which contradicts \( v_3(A/x) \le 1 \).
   - Therefore, \( v_3(A/x) = 0 \) and \( 3^3 \mid x \).
   - This gives \( 3 \prod_{i=3}^n (p_i - a_i + 1) = 3^3 \prod_{i=3}^n p_i^{a_i} \).
   - Since \( a_i \) is odd, \( a_i \ge 1 \), leading to \( p_i^{a_i} \ge p_i \ge p_i - a_i + 1 \), which is a contradiction.

4. **Case II: \( v_2(x) = 1 \)**:
   - This implies \( \frac{A}{x} = 2 \prod_{i=2}^n p_i^{p_i - a_i} \).
   - Therefore, \( d(A/x) = 2 \prod_{i=2}^n (p_i - a_i + 1) = x = 2 \prod_{i=2}^n p_i^{a_i} \).
   - This implies \( \prod_{i=2}^n (p_i - a_i + 1) = \prod_{i=2}^n p_i^{a_i} \).
   - The right-hand side (RHS) is odd, so the left-hand side (LHS) must also be odd.
   - Therefore, \( p_i - a_i + 1 \) is odd for all \( i \ge 2 \), implying \( a_i \) is odd for all \( i \ge 2 \).
   - Since \( a_i \ge 1 \), we have \( p_i^{a_i} \ge p_i \ge p_i - a_i + 1 \) with equality if and only if \( a_i = 1 \).
   - Thus, \( a_i = 1 \) for all \( i \ge 2 \), giving \( x = \prod_{i=1}^n p_i \).

The final answer is \( \boxed{ x = \prod_{i=1}^n p_i } \).