Given a natural number $n\ge 3$, determine all strictly increasing sequences $a_1<a_2<\cdots<a_n$ such that $\text{gcd}(a_1,a_2)=1$ and for any pair of natural numbers $(k,m)$ satisfy $n\ge m\ge 3$, $m\ge k$, $$\frac{a_1+a_2+\cdots +a_m}{a_k}$$ is a positive integer.

1. **Base Case: \( n = 3 \)**

   We start by considering the smallest case where \( n = 3 \). We need to find \( a_1, a_2, a_3 \) such that \( \text{gcd}(a_1, a_2) = 1 \) and for any \( k, m \) with \( 3 \ge m \ge k \ge 1 \), the expression \( \frac{a_1 + a_2 + \cdots + a_m}{a_k} \) is an integer.

   - From the condition \( a_2 \mid a_1 + a_2 + a_3 \), we have:
     \[
     a_2 \mid a_1 + a_2 + a_3 \implies a_2 \mid a_1 + a_3
     \]
   - From the condition \( a_3 \mid a_1 + a_2 + a_3 \), we have:
     \[
     a_3 \mid a_1 + a_2 + a_3 \implies a_3 \mid a_1 + a_2
     \]

   Since \( a_3 \mid a_1 + a_2 \), we can write:
   \[
   a_1 + a_2 = t a_3 \quad \text{for some integer } t
   \]
   Given \( a_1 < a_2 < a_3 \), it follows that:
   \[
   a_1 + a_2 < 2a_3 \implies t = 1 \implies a_3 = a_1 + a_2
   \]

   Now, from \( a_2 \mid 2(a_1 + a_2) \), we get:
   \[
   a_2 \mid 2a_1
   \]
   Since \( \text{gcd}(a_1, a_2) = 1 \), \( a_2 \) must divide 2. Therefore, \( a_2 \) can be 1 or 2. Given \( a_2 > a_1 \ge 1 \), we have:
   \[
   a_1 = 1, \quad a_2 = 2
   \]
   And from \( a_3 = a_1 + a_2 \), we get:
   \[
   a_3 = 3
   \]

2. **Inductive Step: Assume the claim is true for \( n \)**

   Assume that for \( n \), the sequence \( a_1, a_2, \ldots, a_n \) is given by:
   \[
   a_1 = 1, \quad a_2 = 2, \quad a_3 = 3, \quad a_i = 3 \times 2^{i-3} \text{ for } 4 \le i \le n
   \]

3. **Prove for \( n+1 \)**

   Consider the sequence \( a_1, a_2, \ldots, a_{n+1} \) satisfying the conditions. By the induction hypothesis, we have:
   \[
   a_1 = 1, \quad a_2 = 2, \quad a_3 = 3, \quad a_i = 3 \times 2^{i-3} \text{ for } 4 \le i \le n
   \]
   It is easy to see that:
   \[
   a_n = a_1 + a_2 + \cdots + a_{n-1}
   \]

   From the condition \( a_{n+1} \mid a_1 + a_2 + \cdots + a_n \), we have:
   \[
   a_{n+1} \mid 2a_n
   \]
   And from \( a_n \mid a_1 + a_2 + \cdots + a_{n+1} \), we get:
   \[
   a_n \mid 2a_n + a_{n+1} \implies a_n \mid a_{n+1}
   \]
   Since \( a_{n+1} > a_n \), we have:
   \[
   a_{n+1} = 2a_n
   \]

   Therefore, the sequence for \( n+1 \) is:
   \[
   a_1 = 1, \quad a_2 = 2, \quad a_3 = 3, \quad a_i = 3 \times 2^{i-3} \text{ for } 4 \le i \le n+1
   \]

Thus, by induction, the sequence \( a_i \) such that \( a_1 = 1, a_2 = 2, a_3 = 3 \) and for \( 4 \le i \le n \), \( a_i = 3 \times 2^{i-3} \) satisfies the given conditions.

\(\blacksquare\)

The final answer is \( \boxed{ a_i } \) such that \( a_1 = 1, a_2 = 2, a_3 = 3 \) and for \( 4 \le i \le n \), \( a_i = 3 \times 2^{i-3} \).