Find all infinite sequences $a_1, a_2, \ldots$ of positive integers satisfying the following properties:
(a) $a_1 < a_2 < a_3 < \cdots$,
(b) there are no positive integers $i$, $j$, $k$, not necessarily distinct, such that $a_i+a_j=a_k$,
(c) there are infinitely many $k$ such that $a_k = 2k-1$.
1. **Initial Assumption and Setup**:
   - We are given an infinite sequence \(a_1, a_2, \ldots\) of positive integers such that:
     - \(a_1 < a_2 < a_3 < \cdots\)
     - There are no positive integers \(i, j, k\) such that \(a_i + a_j = a_k\)
     - There are infinitely many \(k\) such that \(a_k = 2k - 1\)
   - We need to find all such sequences.

2. **Proving \(a_1 = 1\)**:
   - Assume \(a_1 = c\) where \(c \neq 1\).
   - Pick an index \(k\) such that \(a_k = 2k - 1\).
   - Consider the set \(S = \{1, 2, \ldots, c-1\} \cup \{a_1, a_2, \ldots, a_k\} \cup \{c + a_1, c + a_2, \ldots, c + a_{k-c}\}\).

3. **Analyzing the Set \(S\)**:
   - The sets \(\{1, 2, \ldots, c-1\}\), \(\{a_1, a_2, \ldots, a_k\}\), and \(\{c + a_1, c + a_2, \ldots, c + a_{k-c}\}\) do not intersect each other.
   - Since \(a_i \geq c\) for all \(i \geq 1\), the members of the second and third sets are strictly greater than the members of the first set.
   - The second and third sets do not intersect each other due to the condition \(a_i + a_j \neq a_k\).

4. **Bounding the Elements of \(S\)**:
   - All elements of \(S\) are less than or equal to \(2k - 1\).
   - Since \(a_{k-c} \leq a_k - c = 2k - 1 - c\), all terms \(a_1 + a_i\) for \(i \leq k - c\) satisfy \(a_1 + a_i \leq 2k - 1\).

5. **Contradiction if \(c \neq 1\)**:
   - If \(c \neq 1\), then all members of the second set are at least \(2c\).
   - The numbers \(c, c+1, \ldots, 2c-1\) must be members of the second set, i.e., \(a_1 = c, a_2 = c+1, \ldots, a_{c-1} = 2c-1\).
   - This implies \(a_2 + a_{c-1} = 3c\), which must be in the third set.
   - Therefore, there exists an index \(i \leq k - c\) such that \(a_i = 2c\), leading to a contradiction.

6. **Conclusion from Contradiction**:
   - The assumption \(a_1 = c \neq 1\) leads to a contradiction.
   - Therefore, \(a_1 = 1\).

7. **Inductive Step**:
   - We will prove by induction that \(a_n = 2n - 1\) for each \(n \in \mathbb{N}\).
   - Base case: \(a_1 = 1 = 2 \cdot 1 - 1\).
   - Inductive hypothesis: Assume \(a_i = 2i - 1\) for \(i \leq n\).
   - Consider \(a_{n+1}\). Since \(a_{n+1}\) is the smallest term greater than \(a_n = 2n - 1\), it must be \(2n + 1\).

8. **Verification**:
   - If \(a_{n+1} = 2n + 1\), then \(a_{n+1} = 2(n+1) - 1\).
   - This satisfies the condition \(a_i + a_j \neq a_k\) because the sum of any two odd numbers is even, and all terms in the sequence are odd.

9. **Conclusion**:
   - By induction, \(a_n = 2n - 1\) for all \(n \in \mathbb{N}\).

The final answer is \( \boxed{ a_n = 2n - 1 } \).