(i) Consider two positive integers $a$ and $b$ which are such that $a^a b^b$ is divisible by $2000$. What is the least possible value of $ab$?
(ii) Consider two positive integers $a$ and $b$ which are such that $a^b b^a$ is divisible by $2000$. What is the least possible value of $ab$?
### Part (i)
We need to find the least possible value of \( ab \) such that \( a^a b^b \) is divisible by \( 2000 \). 

1. **Prime Factorization of 2000**:
   \[
   2000 = 2^4 \cdot 5^3
   \]

2. **Divisibility Condition**:
   For \( a^a b^b \) to be divisible by \( 2000 \), it must contain at least \( 2^4 \) and \( 5^3 \) in its prime factorization.

3. **Choosing \( a \) and \( b \)**:
   - We need to ensure that the exponents of 2 and 5 in \( a^a b^b \) are at least 4 and 3, respectively.
   - Introducing other prime factors would only increase the product unnecessarily.

4. **Finding Minimum \( ab \)**:
   - Consider \( a = 1 \) and \( b = 10 \):
     \[
     1^1 \cdot 10^{10} = 10^{10}
     \]
     - \( 10 = 2 \cdot 5 \), so \( 10^{10} = 2^{10} \cdot 5^{10} \).
     - This satisfies the condition since \( 2^{10} \) and \( 5^{10} \) both exceed \( 2^4 \) and \( 5^3 \).

Thus, the least possible value of \( ab \) is:
\[
ab = 1 \cdot 10 = 10
\]

### Part (ii)
We need to find the least possible value of \( ab \) such that \( a^b b^a \) is divisible by \( 2000 \).

1. **Prime Factorization of 2000**:
   \[
   2000 = 2^4 \cdot 5^3
   \]

2. **Divisibility Condition**:
   For \( a^b b^a \) to be divisible by \( 2000 \), it must contain at least \( 2^4 \) and \( 5^3 \) in its prime factorization.

3. **Choosing \( a \) and \( b \)**:
   - Without loss of generality, assume \( 2 \mid a \) and \( 5 \mid b \).
   - We need to ensure that the exponents of 2 and 5 in \( a^b b^a \) are at least 4 and 3, respectively.

4. **Finding Minimum \( ab \)**:
   - Consider \( a = 4 \) and \( b = 5 \):
     \[
     4^5 \cdot 5^4
     \]
     - \( 4 = 2^2 \), so \( 4^5 = (2^2)^5 = 2^{10} \).
     - \( 5^4 = 5^4 \).
     - This satisfies the condition since \( 2^{10} \) and \( 5^4 \) both exceed \( 2^4 \) and \( 5^3 \).

Thus, the least possible value of \( ab \) is:
\[
ab = 4 \cdot 5 = 20
\]

The final answer is \( \boxed{ 10 } \) for part (i) and \( 20 \) for part (ii).