$\boxed{N1}$Let $n$ be a positive integer,$g(n)$ be the number of positive divisors of $n$ of the form $6k+1$ and $h(n)$ be the number of positive divisors of $n$ of the form $6k-1,$where $k$ is a nonnegative integer.Find all positive integers $n$ such that $g(n)$ and $h(n)$ have different parity.
1. Let \( n \) be a positive integer. We define \( g(n) \) as the number of positive divisors of \( n \) of the form \( 6k + 1 \) and \( h(n) \) as the number of positive divisors of \( n \) of the form \( 6k - 1 \), where \( k \) is a nonnegative integer.

2. We start by expressing \( n \) in its prime factorization form:
   \[
   n = 2^x \cdot 3^y \cdot z
   \]
   Here, \( z \) is composed of primes of the form \( 6k - 1 \) or \( 6k + 1 \).

3. Let us further decompose \( z \) into its prime factors:
   \[
   z = c_1^{d_1} \cdot c_2^{d_2} \cdots c_k^{d_k} \cdot e_1^{f_1} \cdot e_2^{f_2} \cdots e_t^{f_t}
   \]
   where \( c_i \) are primes of the form \( 6k - 1 \) and \( e_j \) are primes of the form \( 6k + 1 \).

4. Define:
   \[
   A = (d_1 + 1) \cdot (d_2 + 1) \cdots (d_k + 1)
   \]
   \[
   B = (f_1 + 1) \cdot (f_2 + 1) \cdots (f_t + 1)
   \]
   \( A \) represents the number of ways to choose divisors from the \( c_i \)'s, and \( B \) represents the number of ways to choose divisors from the \( e_j \)'s.

5. Let \( C \) be the number of ways to choose a divisor \( x \) of \( c_1^{d_1} \cdot c_2^{d_2} \cdots c_k^{d_k} \) such that \( x \equiv 1 \pmod{6} \). Then:
   \[
   g(n) = B \cdot C
   \]
   \[
   h(n) = B \cdot (A - C)
   \]

6. We need \( g(n) \) and \( h(n) \) to have different parity. This implies:
   \[
   B \cdot C \not\equiv B \cdot (A - C) \pmod{2}
   \]
   Simplifying, we get:
   \[
   B \cdot C \not\equiv B \cdot A - B \cdot C \pmod{2}
   \]
   \[
   2B \cdot C \not\equiv B \cdot A \pmod{2}
   \]
   Since \( 2B \cdot C \equiv 0 \pmod{2} \), we need:
   \[
   0 \not\equiv B \cdot A \pmod{2}
   \]
   This implies \( B \cdot A \) must be odd.

7. For \( B \cdot A \) to be odd, both \( A \) and \( B \) must be odd. This means all \( d_i \) and \( f_j \) must be even, so that \( d_i + 1 \) and \( f_j + 1 \) are odd.

8. Therefore, \( z \) must be a perfect square, i.e., \( z = t^2 \) for some integer \( t \).

9. Hence, \( n \) must be of the form:
   \[
   n = 2^a \cdot 3^b \cdot t^2
   \]
   where \( a \) and \( b \) are nonnegative integers, and \( t \) is an integer.

The final answer is \( \boxed{ n = 2^a \cdot 3^b \cdot t^2 } \).