How many subsets of $\{1, 2, ... , 2n\}$ do not contain two numbers with sum $2n+1$?
1. **Partition the set**: We start by partitioning the set $\{1, 2, \ldots, 2n\}$ into two subsets:
   \[
   A = \{1, 2, \ldots, n\} \quad \text{and} \quad B = \{n+1, n+2, \ldots, 2n\}
   \]
   This partitioning helps us manage the condition that no two elements in the subset should sum to $2n+1$.

2. **Subset selection from \(A\)**: Let \(S\) be a subset of \(\{1, 2, \ldots, 2n\}\). We need to count the number of ways to select \(k\) elements from \(A\). The number of ways to choose \(k\) elements from \(A\) is given by the binomial coefficient:
   \[
   \binom{n}{k}
   \]
   where \(0 \leq k \leq n\).

3. **Restriction on elements from \(B\)**: For each element \(a_i\) chosen from \(A\), the element \((2n+1) - a_i\) must not be included in \(S\). These elements \((2n+1) - a_1, (2n+1) - a_2, \ldots, (2n+1) - a_k\) are all in \(B\).

4. **Remaining elements in \(B\)**: We can choose the remaining \(n-k\) elements from \(B\) freely. The number of ways to choose these \(n-k\) elements from \(B\) is:
   \[
   2^{n-k}
   \]

5. **Total number of valid subsets for a fixed \(k\)**: The total number of valid subsets \(S\) for a fixed \(k\) is:
   \[
   \binom{n}{k} \cdot 2^{n-k}
   \]

6. **Summing over all possible \(k\)**: To find the total number of valid subsets, we sum over all possible values of \(k\):
   \[
   \sum_{k=0}^{n} \binom{n}{k} 2^{n-k}
   \]

7. **Simplification using the binomial theorem**: We recognize that the sum can be simplified using the binomial theorem. By substituting \(i = n-k\), we get:
   \[
   \sum_{k=0}^{n} \binom{n}{k} 2^{n-k} = \sum_{i=0}^{n} \binom{n}{i} 2^i = (1 + 2)^n = 3^n
   \]

Thus, the total number of subsets of \(\{1, 2, \ldots, 2n\}\) that do not contain two numbers with sum \(2n+1\) is \(3^n\).

The final answer is \(\boxed{3^n}\).