In a board, the positive integer $N$ is written. In each round, Olive can realize any one of the following operations:
I - Switch the current number by a positive multiple of the current number.
II - Switch the current number by a number with the same digits of the current number, but the digits are written in another order(leading zeros are allowed). For instance, if the current number is $2022$, Olive can write any of the following numbers $222,2202,2220$.
Determine all the positive integers $N$, such that, Olive can write the number $1$ after a finite quantity of rounds.
To determine all positive integers \( N \) such that Olive can write the number \( 1 \) after a finite number of rounds, we need to analyze the operations allowed and their implications.

1. **Invariant Analysis for Multiples of 3:**
   - If \( N \) is a multiple of 3, any positive multiple of \( N \) will also be a multiple of 3.
   - Rearranging the digits of a number does not change its residue modulo 3. This is because the sum of the digits of a number determines its residue modulo 3.
   - Since 1 is not a multiple of 3, it is impossible to reach 1 starting from a multiple of 3 using the allowed operations.
   - Therefore, if \( N \) is a multiple of 3, Olive cannot write the number 1.

2. **Algorithm for Non-Multiples of 3:**
   - Suppose \( N \) is not a multiple of 3. We need to show that Olive can reach 1 using the allowed operations.
   - **Step 1: Remove factors of 2 and 5.**
     - If \( N \) has \( 2^a \) and \( 5^b \) in its prime factorization, multiply \( N \) by \( 5^a \) and \( 2^b \). This results in a number that is a multiple of \( 10^{a+b} \).
     - Rearrange the digits to move all zeros to the front, effectively removing the factors of 2 and 5.
   - **Step 2: Handle remaining prime factors.**
     - Now, the number \( N \) has no factors of 2 or 5, and all its prime factors are 7 or greater.
     - Let \( k \) be the current number. By Euler's Theorem, \( k \) divides \( 10^{\varphi(k)} - 1 \), where \( \varphi \) is the Euler's totient function.
     - Find the smallest \( d \) such that \( \gcd(10^d - 1, k) = 1 \).
     - Construct a string \( S_k \) consisting of digits that can be rearranged to form multiples of \( k \).
     - Use Kronecker's Approximation Theorem to find a power of 2 whose leftmost digits match \( S_k \).
     - Rearrange the digits to form a multiple of \( k \), then multiply to get a power of 2.
   - **Step 3: Reduce to 1.**
     - Multiply by appropriate powers of 5 and rearrange digits to move zeros to the front, reducing the number to 1.

3. **Conclusion:**
   - The algorithm works for all non-multiples of 3, allowing Olive to reach 1 after a finite number of rounds.
   - Therefore, the answer is all positive integers \( N \) that are not multiples of 3.

The final answer is \(\boxed{\text{all non-multiples of 3}}\)