Five numbers are chosen from $\{1, 2, \ldots, n\}$. Determine the largest $n$ such that we can always pick some of the 5 chosen numbers so that they can be made into two groups whose numbers have the same sum (a group may contain only one number).
To solve this problem, we need to determine the largest \( n \) such that any five numbers chosen from the set \(\{1, 2, \ldots, n\}\) can always be partitioned into two groups with equal sums.

1. **Understanding the Pigeonhole Principle**:
   The Pigeonhole Principle states that if \( n \) items are put into \( m \) containers, with \( n > m \), then at least one container must contain more than one item. We will use this principle to show that for \( n \leq 15 \), any five numbers chosen from \(\{1, 2, \ldots, n\}\) can always be partitioned into two groups with equal sums.

2. **Sum of Subsets**:
   Consider the set \(\{1, 2, \ldots, n\}\). The number of subsets of any set of size \( n \) is \( 2^n \). For any subset of size 5, there are \( 2^5 = 32 \) possible subsets (including the empty subset). However, we are interested in non-empty subsets, so there are \( 2^5 - 1 = 31 \) non-empty subsets.

3. **Possible Sums**:
   The maximum sum of any subset of 5 numbers from \(\{1, 2, \ldots, n\}\) is achieved by the subset \(\{n-4, n-3, n-2, n-1, n\}\). For \( n = 15 \), this subset is \(\{11, 12, 13, 14, 15\}\), and the sum is \( 11 + 12 + 13 + 14 + 15 = 65 \). Therefore, the possible sums of subsets of 5 numbers from \(\{1, 2, \ldots, 15\}\) range from 1 to 65.

4. **Applying the Pigeonhole Principle**:
   Since there are 31 non-empty subsets and the possible sums range from 1 to 65, by the Pigeonhole Principle, there must be at least two subsets with the same sum if we choose 5 numbers from \(\{1, 2, \ldots, 15\}\). This means that for \( n \leq 15 \), we can always find two subsets with the same sum.

5. **Counterexample for \( n \geq 16 \)**:
   Consider the set \(\{1, 2, 4, 8, 16\}\). The sums of all non-empty subsets are distinct: \( 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31 \). Therefore, for \( n \geq 16 \), it is possible to choose 5 numbers such that no two subsets have the same sum.

Conclusion:
Thus, the largest \( n \) such that any five numbers chosen from \(\{1, 2, \ldots, n\}\) can always be partitioned into two groups with equal sums is \( n = 15 \).

The final answer is \( \boxed{ n = 15 } \).