Determine the smallest integer $n > 1$ with the property that $n^2(n - 1)$ is divisible by 2009.
To determine the smallest integer \( n > 1 \) such that \( n^2(n - 1) \) is divisible by 2009, we first need to factorize 2009.

1. **Factorize 2009:**
   \[
   2009 = 7 \times 7 \times 41 = 7^2 \times 41
   \]
   Therefore, \( n^2(n - 1) \) must be divisible by \( 7^2 \times 41 \).

2. **Divisibility by \( 7^2 \):**
   - For \( n^2 \) to be divisible by \( 7^2 \), \( n \) must be divisible by 7.
   - Let \( n = 7k \) for some integer \( k \).

3. **Divisibility by 41:**
   - We need \( n(n - 1) \) to be divisible by 41.
   - If \( n \) is divisible by 41, then \( n \geq 41 \).
   - If \( n - 1 \) is divisible by 41, then \( n = 41m + 1 \) for some integer \( m \).

4. **Combining conditions:**
   - We need \( n \) to satisfy both conditions: \( n = 7k \) and \( n = 41m + 1 \).
   - This means we need to solve the congruence:
     \[
     7k \equiv 1 \pmod{41}
     \]

5. **Finding the multiplicative inverse of 7 modulo 41:**
   - We need to find \( k \) such that \( 7k \equiv 1 \pmod{41} \).
   - Using the Extended Euclidean Algorithm:
     \[
     41 = 5 \times 7 + 6
     \]
     \[
     7 = 1 \times 6 + 1
     \]
     \[
     6 = 6 \times 1 + 0
     \]
     Back-substituting:
     \[
     1 = 7 - 1 \times (41 - 5 \times 7) = 7 - 41 + 5 \times 7 = 6 \times 7 - 41
     \]
     Thus, the multiplicative inverse of 7 modulo 41 is 6.

6. **Solving for \( n \):**
   - Therefore, \( k = 6 \) and \( n = 7k = 7 \times 6 = 42 \).

7. **Verification:**
   - Check if \( 42^2(42 - 1) \) is divisible by 2009:
     \[
     42^2 = 1764
     \]
     \[
     42 - 1 = 41
     \]
     \[
     1764 \times 41 = 72324
     \]
     \[
     72324 \div 2009 = 36
     \]
     Since 72324 is divisible by 2009, \( n = 42 \) works.

The final answer is \( \boxed{42} \).