We call a set “sum free” if no two elements of the set add up to a third element of the set. What is the maximum size of a sum free subset of $\{ 1, 2, \ldots , 2n - 1 \}$.
To find the maximum size of a sum-free subset of the set $\{1, 2, \ldots, 2n-1\}$, we will use a combination of combinatorial arguments and the pigeonhole principle.

1. **Initial Observation**:
   - The set $\{1, 2, \ldots, 2n-1\}$ contains $2n-1$ elements.
   - We need to find a subset $A$ such that no two elements in $A$ add up to another element in $A$.

2. **Sum-Free Subset Size**:
   - We claim that the maximum size of a sum-free subset of $\{1, 2, \ldots, 2n-1\}$ is $n$.

3. **Proof by Contradiction**:
   - Assume there exists a sum-free subset $A$ of size $m$ where $m > n$.
   - We will show that this leads to a contradiction.

4. **Grouping Elements**:
   - Consider the following groupings of the elements of $\{1, 2, \ldots, 2n-1\}$:
     \[
     \{1, 2n-1\}, \{2, 2n-2\}, \ldots, \{n-1, n+1\}, \{n\}
     \]
   - This grouping has $n$ pairs and one single element, $n$.

5. **Pigeonhole Principle**:
   - Since $m > n$, by the pigeonhole principle, at least one of these pairs must have both elements in $A$.
   - If both elements of any pair $\{k, 2n-k\}$ are in $A$, then $k + (2n-k) = 2n$ is in $A$, which is a contradiction because $2n$ is not in the set $\{1, 2, \ldots, 2n-1\}$.

6. **Alternative Grouping**:
   - Consider another grouping:
     \[
     \{2, 2n-1\}, \{3, 2n-2\}, \ldots, \{n, n+1\}, \{1\}
     \]
   - This grouping also has $n$ pairs and one single element, $1$.

7. **Analysis of Groupings**:
   - In both groupings, there can be at most one element from each pair in $A$ to avoid the sum being another element in $A$.
   - Therefore, the maximum number of elements in $A$ is $n$.

8. **Conclusion**:
   - If $m > n$, then $A$ cannot be sum-free, leading to a contradiction.
   - Hence, the maximum size of a sum-free subset of $\{1, 2, \ldots, 2n-1\}$ is $n$.

The final answer is $\boxed{n}$