Amy and Bob choose numbers from $0,1,2,\cdots,81$ in turn and Amy choose the number first. Every time the one who choose number chooses one number from the remaining numbers. When all $82$ numbers are chosen, let $A$ be the sum of all the numbers Amy chooses, and let $B$ be the sum of all the numbers Bob chooses. During the process, Amy tries to make $\gcd(A,B)$ as great as possible, and Bob tries to make $\gcd(A,B)$ as little as possible. Suppose Amy and Bob take the best strategy of each one, respectively, determine $\gcd(A,B)$ when all $82$ numbers are chosen.
1. **Sum of All Numbers**:
   The sum of all numbers from \(0\) to \(81\) is given by the formula for the sum of an arithmetic series:
   \[
   S = \sum_{k=0}^{81} k = \frac{81 \cdot (81 + 1)}{2} = \frac{81 \cdot 82}{2} = 81 \cdot 41
   \]
   Therefore, \(A + B = 81 \cdot 41\).

2. **Applying the Euclidean Algorithm**:
   Using the Euclidean Algorithm, we have:
   \[
   \gcd(A, B) = \gcd(A, A + B) = \gcd(A, 81 \cdot 41)
   \]
   Since \(81 = 3^4\), we get:
   \[
   \gcd(A, 81 \cdot 41) = \gcd(A, 3^4 \cdot 41)
   \]
   This implies that \(\gcd(A, B)\) must be a divisor of \(3^4 \cdot 41\).

3. **Bob's Strategy to Minimize \(\gcd(A, B)\)**:
   Bob aims to make \(\gcd(A, B)\) as small as possible. To achieve this, Bob can ensure that neither \(A\) nor \(B\) is divisible by \(3\). 

   - **Dividing Numbers into Sets**:
     Divide the numbers into 14 sets:
     \[
     A_1 = \{0, 1, 2, 3\}, \quad A_2 = \{4, 5, 6, 7, 8, 9\}, \quad \dots, \quad A_{14} = \{76, 77, 78, 79, 80, 81\}
     \]
     Each set \(A_i\) for \(i > 1\) contains 6 consecutive numbers.

   - **Ensuring Non-Divisibility by 3**:
     For \(A_1\), if Amy takes a number divisible by 3, Bob takes another number divisible by 3. This ensures that the sums of numbers taken from \(A_1\) by both Amy and Bob are not divisible by 3.
     For \(A_i\) (\(i \geq 2\)), if Amy takes a number of the form \(3k + l\), Bob takes another number of the same form. This ensures that the sums of numbers taken from each \(A_i\) by both Amy and Bob are divisible by 3. Summing these sets ensures that \(A\) and \(B\) are not divisible by 3.

4. **Amy's Strategy to Maximize \(\gcd(A, B)\)**:
   Amy aims to make \(\gcd(A, B)\) as large as possible. To achieve this, Amy can ensure that both \(A\) and \(B\) are divisible by 41.

   - **Ensuring Divisibility by 41**:
     Write the 82 numbers as two sets of complete residues modulo 41. Amy takes the first number randomly. If Bob takes the same number, Amy takes another random number. This procedure continues until Bob takes a different number from Amy. Then Amy takes the same number Bob took, and this process repeats. This ensures that both Amy and Bob will take each residue exactly once.
     \[
     A \equiv B \equiv 0 + 1 + 2 + \cdots + 40 \equiv 41 \cdot 20 \equiv 0 \pmod{41}
     \]
     Therefore, both \(A\) and \(B\) are divisible by 41.

5. **Conclusion**:
   Since Bob ensures that \(A\) and \(B\) are not divisible by 3, and Amy ensures that both \(A\) and \(B\) are divisible by 41, the greatest common divisor of \(A\) and \(B\) is 41.

\(\blacksquare\)

The final answer is \( \boxed{ \gcd(A, B) = 41 } \).