Find all integers $ n\ge 2$ having the following property: for any $ k$ integers $ a_{1},a_{2},\cdots,a_{k}$ which aren't congruent to each other (modulo $ n$), there exists an integer polynomial $ f(x)$ such that congruence equation $ f(x)\equiv 0 (mod n)$ exactly has $ k$ roots $ x\equiv a_{1},a_{2},\cdots,a_{k} (mod n).$
To solve the problem, we need to find all integers \( n \geq 2 \) such that for any \( k \) integers \( a_1, a_2, \ldots, a_k \) which are not congruent to each other modulo \( n \), there exists an integer polynomial \( f(x) \) such that the congruence equation \( f(x) \equiv 0 \pmod{n} \) has exactly \( k \) roots \( x \equiv a_1, a_2, \ldots, a_k \pmod{n} \).

1. **Step 1: Show that \( n \) must be a power of a prime.**

   Suppose \( n \) is not a power of a prime. Then \( n \) can be written in the form \( n = pq \) where \( p \) and \( q \) are relatively prime integers greater than 1. Consider the \( n-1 \) numbers \( \{1, 2, \ldots, n-1\} \). By the condition of the problem, there exists a polynomial \( f \) such that the equation \( f(x) \equiv 0 \pmod{n} \) has exactly \( n-1 \) solutions \( 1, 2, \ldots, n-1 \). We will show that 0 is also a solution of this equation.

   Since \( p, q > 1 \), we have \( p, q < n-1 \). By the fundamental property of polynomials, we have:
   \[
   p \mid f(p) - f(0) \quad \text{and} \quad q \mid f(q) - f(0)
   \]
   Since \( p \mid f(p) \) and \( q \mid f(q) \), it follows that:
   \[
   pq \mid f(0)
   \]
   This implies \( f(0) \equiv 0 \pmod{n} \), which contradicts the assumption that \( f(x) \equiv 0 \pmod{n} \) has exactly \( n-1 \) solutions. Therefore, \( n \) must be a power of a prime.

2. **Step 2: Consider \( n = p^a \) where \( p \) is a prime.**

   We have two cases to consider: \( p \) is an odd prime and \( p = 2 \).

   - **Case 1: \( p \) is an odd prime.**

     We will show that \( a = 1 \). Suppose \( a > 1 \). Consider the two numbers \( 0 \) and \( p^{a-1} \). By the condition of the problem, there exists a polynomial \( f \) such that the equation \( f(x) \equiv 0 \pmod{p^a} \) has exactly two solutions \( 0 \) and \( p^{a-1} \). We can write \( f \) in the form:
     \[
     f(x) = (x - 0)(x - p^{a-1})G(x) + ax + b
     \]
     Since \( p^a \mid f(0) \) and \( p^a \mid f(p^{a-1}) \), it follows that \( p^a \mid b \) and \( p \mid a \). Consider \( x = 2p^{a-1} \). It is not hard to show that \( x = 2p^{a-1} \) is a solution of the equation \( f(x) \equiv 0 \pmod{p^a} \), which contradicts the assumption that \( f(x) \equiv 0 \pmod{p^a} \) has exactly two solutions. Hence, \( a = 1 \), and \( n = p \) is a prime number that satisfies the condition.

   - **Case 2: \( p = 2 \).**

     By a similar argument, we have \( a \leq 3 \) (by considering the two numbers \( 0 \) and \( 2^{a-2} \) and another number \( 2^{a-1} \)). If \( a = 3 \), we only need to consider the two numbers \( 0 \), \( 6 \), and \( 4 \). Hence, \( a \leq 2 \), and then \( n \in \{2, 4\} \).

Thus, all numbers that satisfy the condition are \( n = 2, 4, p \) where \( p \) is a prime number.

The final answer is \( \boxed{ n = 2, 4, p } \) where \( p \) is a prime number.