A finite set of integers is called [i]bad[/i] if its elements add up to $2010$. A finite set of integers is a [i]Benelux-set[/i] if none of its subsets is bad. Determine the smallest positive integer $n$ such that the set $\{502, 503, 504, . . . , 2009\}$ can be partitioned into $n$ Benelux-sets.
(A partition of a set $S$ into $n$ subsets is a collection of $n$ pairwise disjoint subsets of $S$, the union of which equals $S$.)

[i](2nd Benelux Mathematical Olympiad 2010, Problem 1)[/i]
1. **Understanding the Problem:**
   We need to partition the set \( \{502, 503, 504, \ldots, 2009\} \) into the smallest number \( n \) of subsets such that none of the subsets contains a subset whose elements sum to 2010. These subsets are called Benelux-sets.

2. **Initial Observations:**
   - A set with a single element cannot sum to 2010.
   - A set with two elements can sum to 2010 if the elements are \( a \) and \( b \) such that \( a + b = 2010 \).
   - A set with three elements can sum to 2010 if the elements are \( a, b, \) and \( c \) such that \( a + b + c = 2010 \).

3. **Bounding the Elements:**
   - The smallest element in the set is 502.
   - The largest element in the set is 2009.
   - If a set contains four or more elements, the minimum possible sum is \( 502 + 503 + 504 + 505 = 2014 \), which is greater than 2010. Therefore, a bad set can only have 2 or 3 elements.

4. **Identifying Bad Pairs and Triplets:**
   - Bad pairs: \( (a, b) \) such that \( a + b = 2010 \).
   - Bad triplets: \( (a, b, c) \) such that \( a + b + c = 2010 \).

5. **Partitioning the Set:**
   - Consider the pairs \( (502, 1508), (503, 1507), (504, 1506), \ldots, (1004, 1006) \). These pairs sum to 2010 and must be in different subsets.
   - We can split the set into two subsets:
     \[
     A = \{502, 503, 504, \ldots, 670, 1006, 1007, 1008, \ldots, 1339\}
     \]
     \[
     B = \{671, 672, 673, \ldots, 1004, 1005, 1340, 1341, 1342, \ldots, 1508\}
     \]

6. **Verifying the Partition:**
   - In subset \( A \), the maximum sum of any three elements is \( 668 + 669 + 670 = 2007 \), which is less than 2010. Therefore, no bad triplets exist in \( A \).
   - In subset \( B \), the minimum sum of any three elements is \( 671 + 672 + 673 = 2016 \), which is greater than 2010. Therefore, no bad triplets exist in \( B \).

7. **Conclusion:**
   - We have shown that the set \( \{502, 503, 504, \ldots, 2009\} \) can be partitioned into 2 subsets, each of which is a Benelux-set.

The final answer is \( \boxed{ n = 2 } \).