Find all $k>0$ for which a strictly decreasing function $g:(0;+\infty)\to(0;+\infty)$ exists such that $g(x)\geq kg(x+g(x))$ for all positive $x$.
To find all \( k > 0 \) for which a strictly decreasing function \( g: (0, +\infty) \to (0, +\infty) \) exists such that \( g(x) \geq k g(x + g(x)) \) for all positive \( x \), we will analyze the given inequality and the behavior of the function \( g \).

1. **Case \( k \leq 1 \):**
   - If \( k \leq 1 \), then for any \( x > 0 \), we have:
     \[
     k g(x + g(x)) \leq g(x + g(x)) \leq g(x)
     \]
     This inequality holds because \( g \) is strictly decreasing, meaning \( g(x + g(x)) \leq g(x) \). Therefore, the condition \( g(x) \geq k g(x + g(x)) \) is satisfied for all \( x > 0 \).

2. **Case \( k > 1 \):**
   - Suppose \( k > 1 \). We will show that this leads to a contradiction.
   - Assume \( g(x) \leq \frac{x}{k} \) for some \( x > 0 \). Then:
     \[
     \frac{x}{k} \geq g(x) \geq k g(x + g(x)) \geq k g\left(x + \frac{x}{k}\right)
     \]
     Since \( g \) is strictly decreasing, we have:
     \[
     g\left(x + \frac{x}{k}\right) \leq g(x)
     \]
     Therefore:
     \[
     k g\left(x + \frac{x}{k}\right) \leq k g(x)
     \]
     This implies:
     \[
     \frac{x}{k} \geq k g\left(x + \frac{x}{k}\right) \Rightarrow g\left(x + \frac{x}{k}\right) \leq \frac{x}{k^2}
     \]
   - Now, let \( x = 2x \):
     \[
     g(2x) \leq \frac{x}{k^2}
     \]
   - Repeating this process, we get:
     \[
     g(x) \leq \frac{x}{nk^n}
     \]
     for any \( n \). As \( n \to \infty \), \( \frac{n+1}{n} \to 1 \), so:
     \[
     g(x) \geq k g\left(\frac{n+1}{n} x\right)
     \]
     Since \( g \) is strictly decreasing, this implies:
     \[
     g(x) \geq k g(x)
     \]
     which is a contradiction because \( k > 1 \).

Therefore, the only possible values for \( k \) are those where \( k \leq 1 \).

The final answer is \( \boxed{ k \leq 1 } \).