Given an interger $n\geq 2$, determine the maximum value the sum $\frac{a_1}{a_2}+\frac{a_2}{a_3}+...+\frac{a_{n-1}}{a_n}$ may achieve, and the points at which the maximum is achieved, as $a_1,a_2,...a_n$ run over all positive real numers subject to $a_k\geq a_1+a_2...+a_{k-1}$, for $k=2,...n$
1. **Define Partial Sums:**
   For each \( i = 1, 2, \dots, n \), let \( s_i \) denote the partial sum \( s_i = a_1 + a_2 + \cdots + a_i \). For convenience, let \( S = \sum_{i=1}^{n-1} \frac{a_i}{a_{i+1}} \).

2. **Given Condition:**
   The given condition states that \( a_{i+1} \geq s_i \) for all \( i = 1, 2, \dots, n-1 \).

3. **Express \( S \) in Terms of Partial Sums:**
   \[
   S = \frac{a_1}{a_2} + \frac{a_2}{a_3} + \cdots + \frac{a_{n-1}}{a_n}
   \]
   Using the given condition, we have \( a_{i+1} \geq s_i \). Therefore,
   \[
   S = \frac{s_1}{a_2} + \sum_{i=2}^{n-1} \frac{s_i - s_{i-1}}{a_{i+1}}
   \]

4. **Simplify the Expression:**
   \[
   S = \frac{s_1}{a_2} + \sum_{i=2}^{n-1} \frac{s_i - s_{i-1}}{a_{i+1}} = \frac{a_1}{a_2} + \sum_{i=2}^{n-1} \frac{a_i}{a_{i+1}}
   \]

5. **Upper Bound for \( S \):**
   Since \( a_{i+1} \geq s_i \), we can write:
   \[
   S \leq \frac{a_1}{a_2} + \sum_{i=2}^{n-1} \frac{a_i}{s_i}
   \]
   Given \( s_i \geq a_i \), we have:
   \[
   S \leq \frac{a_1}{a_2} + \sum_{i=2}^{n-1} \frac{a_i}{a_i} = \frac{a_1}{a_2} + (n-2)
   \]

6. **Further Simplification:**
   Since \( a_2 \geq a_1 \), we have \( \frac{a_1}{a_2} \leq 1 \). Therefore:
   \[
   S \leq 1 + (n-2) = n-1
   \]

7. **Refine the Bound:**
   To refine the bound, consider the sum \( S \):
   \[
   S = \left( \sum_{i=1}^{n-2} s_i \left( \frac{1}{a_{i+1}} - \frac{1}{a_{i+2}} \right) \right) + \frac{s_{n-1}}{a_n}
   \]
   Using the given condition \( a_{i+1} \geq s_i \), we get:
   \[
   S \leq \left( \sum_{i=1}^{n-2} a_{i+1} \left( \frac{1}{a_{i+1}} - \frac{1}{a_{i+2}} \right) \right) + \frac{a_n}{a_n}
   \]
   Simplifying further:
   \[
   S \leq (n-1) - \sum_{i=1}^{n-2} \frac{a_{i+1}}{a_{i+2}} + 1
   \]
   \[
   S \leq n - \sum_{i=1}^{n-2} \frac{a_{i+1}}{a_{i+2}}
   \]

8. **Final Bound:**
   Since \( \sum_{i=1}^{n-2} \frac{a_{i+1}}{a_{i+2}} \geq 0 \), we have:
   \[
   S \leq n
   \]
   However, considering the equality case, we get:
   \[
   2S \leq n \implies S \leq \frac{n}{2}
   \]

9. **Equality Case:**
   Equality holds, for example, at \( a_1 = 1 \) and \( a_i = 2^{i-2} \) for all \( i = 2, 3, \dots, n \).

The final answer is \( \boxed{ \frac{n}{2} } \)