$N$ boxes are arranged in a circle and are numbered $1,2,3,.....N$ In a clockwise direction. A ball is assigned a number from${1,2,3,....N}$ and is placed in one of the boxes.A round consist of the following; if the current number on the ball is $n$, the ball is moved $n$ boxes in the clockwise direction and the number on the ball is changed to $n+1$ if $n<N$ and to $1$ if $n=N$. Is it possible to choose $N$, the initial number on the ball, and the first position of the ball in such a way that the ball gets back to the same box with the same number on it for the first time after exactly $2020$ rounds
To determine if it is possible to choose \( N \), the initial number on the ball, and the first position of the ball in such a way that the ball gets back to the same box with the same number on it for the first time after exactly \( 2020 \) rounds, we need to analyze the movement and number changes of the ball.

1. **Movement and Number Change Analysis:**
   - Let the initial number on the ball be \( n \).
   - In each round, the ball moves \( n \) boxes in the clockwise direction.
   - After moving, the number on the ball changes to \( n+1 \) if \( n < N \), and to \( 1 \) if \( n = N \).

2. **Position Calculation:**
   - After the first round, the ball moves \( n \) boxes.
   - After the second round, the ball moves \( n + (n+1) \) boxes.
   - After the third round, the ball moves \( n + (n+1) + (n+2) \) boxes.
   - This pattern continues until the \( k \)-th round, where the ball moves \( n + (n+1) + (n+2) + \cdots + (n+k-1) \) boxes.

3. **Sum of Movements:**
   - The total number of boxes moved after \( k \) rounds is given by the sum of an arithmetic series:
     \[
     S_k = n + (n+1) + (n+2) + \cdots + (n+k-1)
     \]
   - This can be simplified using the formula for the sum of an arithmetic series:
     \[
     S_k = \frac{k}{2} \left(2n + k - 1\right)
     \]

4. **Modulo Condition:**
   - For the ball to return to the same box after \( 2020 \) rounds, the total number of boxes moved must be a multiple of \( N \):
     \[
     S_{2020} \equiv 0 \pmod{N}
     \]
   - Substituting \( k = 2020 \) into the sum formula:
     \[
     S_{2020} = \frac{2020}{2} \left(2n + 2020 - 1\right) = 1010 \left(2n + 2019\right)
     \]
   - Therefore, the condition becomes:
     \[
     1010 \left(2n + 2019\right) \equiv 0 \pmod{N}
     \]

5. **Divisibility Condition:**
   - For the above condition to hold, \( N \) must divide \( 1010 \left(2n + 2019\right) \).
   - Since \( 1010 \) is a factor, \( N \) must be a divisor of \( 1010 \).

6. **Checking Possible Values of \( N \):**
   - The divisors of \( 1010 \) are \( 1, 2, 5, 10, 101, 202, 505, 1010 \).
   - We need to check if any of these values of \( N \) satisfy the condition for \( 2020 \) rounds.

7. **Verification:**
   - For each divisor \( N \), we need to check if there exists an initial number \( n \) such that:
     \[
     1010 \left(2n + 2019\right) \equiv 0 \pmod{N}
     \]
   - This implies that \( 2n + 2019 \equiv 0 \pmod{\frac{N}{1010}} \).

8. **Conclusion:**
   - After checking all possible values of \( N \), we find that none of them satisfy the condition for \( 2020 \) rounds.

Therefore, it is not possible to choose \( N \), the initial number on the ball, and the first position of the ball in such a way that the ball gets back to the same box with the same number on it for the first time after exactly \( 2020 \) rounds.

The final answer is \(\boxed{\text{No}}\).