Determine all non empty sets $C_1, C_2, C_3, \cdots $ such that each one of them has a finite number of elements, all their elements are positive integers, and they satisfy the following property: For any positive integers $n$ and $m$, the number of elements in the set $C_n$ plus the number of elements in the set $C_m$ equals the sum of the elements in the set $C_{m + n}$.

[i]Note:[/i] We denote $\lvert C_n \lvert$ the number of elements in the set $C_n$, and $S_k$ as the sum of the elements in the set $C_n$ so the problem's condition is that for every $n$ and $m$:
\[\lvert C_n \lvert + \lvert C_m \lvert = S_{n + m}\]
is satisfied.
1. Let \( k \) be the minimal element of \( \{|C_1|, |C_2|, \ldots\} \) and let \( C_j \) be a set which achieves this size. Then we must have:
   \[
   2k = |C_j| + |C_j| = S_{2j} = |C_1| + |C_{2j-1}| \ge k + k
   \]
   Since \( k \) is the minimal size, all equalities must hold, so \( |C_1| = k \).

2. Since \( |C_2| \ge k \), we must have:
   \[
   S_2 \ge 1 + 2 + \cdots + k = \frac{k(k+1)}{2}
   \]
   Using the given condition \( 2k = |C_1| + |C_1| = S_2 \), we get:
   \[
   2k \ge \frac{k(k+1)}{2}
   \]
   Simplifying this inequality:
   \[
   4k \ge k(k+1) \implies 4 \ge k + 1 \implies 3 \ge k
   \]
   Therefore, \( k \) can be 1, 2, or 3.

3. Let's consider the sets \( A_1 = \{2\}, A_2 = \{1, 3\}, A_3 = \{1, 2, 3\} \) and note that \( |A_i| = i \).

4. **Lemma:** If \( S_i = 2k \), then \( C_i = A_k \).

   **Proof:**
   - If \( k = 1 \), then \( C_i \) cannot have more than one element, otherwise \( S_i \ge 1 + 2 > 2 \). So \( C_i \) has one element which must be 2.
   - If \( k = 2 \), then \( C_i \) cannot have more than two elements, otherwise \( S_i \ge 1 + 2 + 3 > 4 \). So \( C_i \) has exactly two elements, and since \( 1 + 3 \) is the only way to represent 4 as a sum of two distinct elements, \( C_i = A_2 \).
   - If \( k = 3 \), \( S_i \ge 1 + 2 + 3 \) and since equality is achieved, we must have exactly \( C_i = A_3 \).

   This finishes the proof. \(\blacksquare\)

5. Then we must have \( A_k = C_2 = C_3 = \cdots \) by induction. It is clear that the sequences that satisfy \( |C_1| = k \le 3 \) and \( A_k = C_2 = C_3 = \cdots \) all satisfy the problem condition, so these are all the possibilities.

The final answer is \( \boxed{ \{A_1, A_2, A_3\} } \) where \( A_1 = \{2\}, A_2 = \{1, 3\}, A_3 = \{1, 2, 3\} \).