Does there exist a set $A\subseteq\mathbb{N}^*$ such that for any positive integer $n$, $A\cap\{n,2n,3n,...,15n\}$ contains exactly one element and there exists infinitely many positive integer $m$ such that $\{m,m+2018\}\subset A$? Please prove your conclusion.
To determine whether there exists a set \( A \subseteq \mathbb{N}^* \) such that for any positive integer \( n \), \( A \cap \{n, 2n, 3n, \ldots, 15n\} \) contains exactly one element and there exist infinitely many positive integers \( m \) such that \( \{m, m+2018\} \subset A \), we can proceed as follows:

1. **Define the Set \( A \)**:
   Consider the set \( A \) defined by:
   \[
   A = \{ n \in \mathbb{Z}^+ \mid f(n) \equiv 0 \pmod{15} \}
   \]
   where \( f(n) = \nu_2(n) + 4\nu_3(n) + 9\nu_5(n) + 11\nu_7(n) + 7\nu_{11}(n) + 14\nu_{13}(n) \).

2. **Properties of \( f(n) \)**:
   Here, \( \nu_p(n) \) denotes the highest power of the prime \( p \) that divides \( n \). The function \( f(n) \) is constructed such that it takes into account the exponents of the prime factors of \( n \) modulo 15.

3. **Uniqueness in \( \{n, 2n, \ldots, 15n\} \)**:
   For any positive integer \( n \), consider the set \( \{n, 2n, 3n, \ldots, 15n\} \). We need to show that \( A \cap \{n, 2n, 3n, \ldots, 15n\} \) contains exactly one element. Since \( f(n) \equiv 0 \pmod{15} \) and \( f(n) \) is defined in terms of the exponents of the prime factors of \( n \), the function \( f \) ensures that exactly one of \( n, 2n, \ldots, 15n \) will satisfy \( f(k) \equiv 0 \pmod{15} \).

4. **Existence of \( m \) and \( m+2018 \) in \( A \)**:
   We need to show that there exist infinitely many positive integers \( m \) such that \( \{m, m+2018\} \subset A \). This means both \( m \) and \( m+2018 \) must satisfy \( f(m) \equiv 0 \pmod{15} \) and \( f(m+2018) \equiv 0 \pmod{15} \).

5. **Choosing \( m \)**:
   Consider \( m \) such that both \( m \) and \( m+2018 \) are not divisible by any of the primes 2, 3, 5, 7, 11, and 13. This ensures that the exponents \( \nu_p(m) \) and \( \nu_p(m+2018) \) are zero for these primes. Consequently, \( f(m) \equiv 0 \pmod{15} \) and \( f(m+2018) \equiv 0 \pmod{15} \).

6. **Infinitely Many \( m \)**:
   Since there are infinitely many integers that are not divisible by any of the primes 2, 3, 5, 7, 11, and 13, there are infinitely many such \( m \) satisfying the condition \( \{m, m+2018\} \subset A \).

Conclusion:
Thus, the set \( A \) defined as \( A = \{ n \in \mathbb{Z}^+ \mid f(n) \equiv 0 \pmod{15} \} \) satisfies the given conditions.

The final answer is \( \boxed{ A = \{ n \in \mathbb{Z}^+ \mid f(n) \equiv 0 \pmod{15} \} } \).