Determine the least $n\in\mathbb{N}$ such that $n!=1\cdot 2\cdot 3\cdots (n-1)\cdot n$ has at least $2010$ positive factors.
To determine the least \( n \in \mathbb{N} \) such that \( n! = 1 \cdot 2 \cdot 3 \cdots (n-1) \cdot n \) has at least \( 2010 \) positive factors, we need to use the properties of the prime factorization of \( n! \).

1. **Prime Factorization of \( n! \)**:
   If the prime factorization of \( N \) is \( N = p_1^{a_1} \cdot p_2^{a_2} \cdots p_k^{a_k} \) for distinct primes \( p_1, p_2, \ldots, p_k \), then the number of factors of \( N \) is given by:
   \[
   (a_1 + 1)(a_2 + 1) \cdots (a_k + 1)
   \]

2. **Prime Factorization of \( 13! \)**:
   \[
   13! = 2^{10} \cdot 3^5 \cdot 5^2 \cdot 7 \cdot 11 \cdot 13
   \]
   The number of factors of \( 13! \) is:
   \[
   (10 + 1)(5 + 1)(2 + 1)(1 + 1)(1 + 1)(1 + 1) = 11 \cdot 6 \cdot 3 \cdot 2 \cdot 2 \cdot 2 = 1584
   \]
   Since \( 1584 < 2010 \), \( 13! \) does not have enough factors.

3. **Prime Factorization of \( 14! \)**:
   \[
   14! = 2^{11} \cdot 3^5 \cdot 5^2 \cdot 7^2 \cdot 11 \cdot 13
   \]
   The number of factors of \( 14! \) is:
   \[
   (11 + 1)(5 + 1)(2 + 1)(2 + 1)(1 + 1)(1 + 1) = 12 \cdot 6 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 2592
   \]
   Since \( 2592 \geq 2010 \), \( 14! \) has enough factors.

Therefore, the least \( n \) such that \( n! \) has at least \( 2010 \) positive factors is \( n = 14 \).

The final answer is \( \boxed{14} \).