Let $n{}$ be a positive integer, and let $\mathcal{C}$ be a collection of subsets of $\{1,2,\ldots,2^n\}$ satisfying both of the following conditions:[list=1]
[*]Every $(2^n-1)$-element subset of $\{1,2,\ldots,2^n\}$ is a member of $\mathcal{C}$, and
[*]Every non-empty member $C$ of $\mathcal{C}$ contains an element $c$ such that $C\setminus\{c\}$ is again a member of $\mathcal{C}$.
[/list]Determine the smallest size $\mathcal{C}$ may have.

[i]Serbia, Pavle Martinovic ́[/i]
1. **Base Case for \( n = 1 \)**:
   - For \( n = 1 \), the set \( \{1, 2\} \) has subsets \( \emptyset, \{1\}, \{2\}, \{1, 2\} \).
   - The collection \( \mathcal{C}_1 = \{\emptyset, \{1\}, \{2\}\} \) satisfies the conditions:
     - Every \( (2^1 - 1) = 1 \)-element subset is in \( \mathcal{C}_1 \).
     - Every non-empty subset contains an element whose removal leaves another subset in \( \mathcal{C}_1 \).

2. **Inductive Step**:
   - Assume for \( n-1 \), there exists a collection \( \mathcal{C}_{n-1} \) of size \( n \cdot 2^{n-1} + 1 \) satisfying the conditions.
   - Construct \( \mathcal{C}_n \) for \( n \) as follows:
     - Let \( X = \{1, 2, \ldots, 2^n\} \).
     - Form \( \mathcal{C}_{n-1}' \) by taking all sets of the form \( C \cup \{2^{n-1}+1, 2^{n-1}+2, \ldots, 2^n\} \) for \( C \in \mathcal{C}_{n-1} \).
     - Form \( \mathcal{C}_{n-1}'' \) by taking all sets of the form \( \{2^n + 1 - c : c \in C\} \) for \( C \in \mathcal{C}_{n-1}' \).
     - Combine \( \mathcal{C}_{n-1}' \cup \mathcal{C}_{n-1}'' \) and add the empty set and the \( 2n-2 \) sets:
       \[
       \{1\}, \{2^{n-1}+1\}, \{1, 2\}, \{2^{n-1}+1, 2^{n-1}+2\}, \ldots, \{1, 2, \ldots, 2^{n-1}-1\}, \{2^{n-1}+1, 2^{n-1}+2, \ldots, 2^n-1\}
       \]
     - This collection \( \mathcal{C}_n \) has size \( n \cdot 2^n + 1 \) and satisfies the conditions.

3. **Lower Bound Proof**:
   - Let \( X = \{1, 2, \ldots, 2^n\} \) and \( \mathcal{C} \) be a collection satisfying the conditions.
   - Assume \( \mathcal{C} \) contains the empty set and not \( X \).
   - For each non-empty \( C \in \mathcal{C} \), choose an element \( x_C \) such that \( C \setminus \{x_C\} \in \mathcal{C} \).
   - Remove all maximal members of size less than \( |X| - 1 \) until the collection has no more such members.
   - The resulting collection has maximal members of size \( |X| - 1 \).

4. **Tree Structure and Lemma**:
   - Define a tree on vertex set \( \mathcal{C} \) with each non-empty \( C \) having parent \( C \setminus \{x_C\} \).
   - For each vertex \( C \), let \( h_C \) be the distance to the nearest leaf, \( s_C \) the number of leaves in the subtree rooted at \( C \), and \( v_C \) the size of the vertex set of this subtree.
   - **Lemma**: \( h_C \geq s_C \) and \( v_C \geq s_C \log_2 s_C + h_C - s_C + 1 \).

5. **Proof of Lemma**:
   - Base case: \( C \) is a leaf, \( h_C = 1 \), \( s_C = 1 \), and the inequality holds.
   - Induction step: Let \( C_1, \ldots, C_m \) be the children of \( C \).
     - \( h_{C_i} = h_C - 1 \), \( s_{C_1} + \cdots + s_{C_m} = s_C \), and \( v_{C_1} + \cdots + v_{C_m} = v_C - 1 \).
     - Using induction hypothesis:
       \[
       v_C = 1 + \sum_{i=1}^m v_{C_i} \geq 1 + \sum_{i=1}^m (s_{C_i} \log_2 s_{C_i} + h_{C_i} - s_{C_i} + 1)
       \]
       \[
       = \sum_{i=1}^m s_{C_i} \log_2 s_{C_i} + \sum_{i=1}^m (h_C - 1) - s_C + m + 1
       \]
       \[
       = \sum_{i=1}^m s_{C_i} \log_2 s_{C_i} + m h_C - s_C + 1
       \]
       \[
       \geq \sum_{i=1}^m s_{C_i} \log_2 s_{C_i} + (m-1) s_C + h_C - s_C + 1
       \]
       \[
       \geq s_C \log_2 s_C + h_C - s_C + 1
       \]
     - Using convexity of \( x \log_2 x \):
       \[
       \sum_{i=1}^m s_{C_i} \log_2 s_{C_i} \geq s_C \log_2 \left( \frac{s_C}{m} \right)
       \]
       \[
       \log_2 \left( \frac{s_C}{m} \right) + m - 1 \geq \log_2 s_C
       \]
       \[
       \sum_{i=1}^m s_{C_i} \log_2 s_{C_i} + (m-1) s_C \geq s_C \log_2 s_C
       \]

6. **Conclusion**:
   - The smallest size \( \mathcal{C} \) may have is \( n \cdot 2^n + 1 \).

The final answer is \( \boxed{ n \cdot 2^n + 1 } \).