Let $n\geq 2$ be a positive integer and let $a_1,a_2,...,a_n\in[0,1]$ be real numbers. Find the maximum value of the smallest of the numbers: \[a_1-a_1a_2, \ a_2-a_2a_3,...,a_n-a_na_1.\]
1. **Define the problem and variables:**
   Let \( n \geq 2 \) be a positive integer and let \( a_1, a_2, \ldots, a_n \in [0,1] \) be real numbers. We need to find the maximum value of the smallest of the numbers:
   \[
   a_1 - a_1 a_2, \ a_2 - a_2 a_3, \ \ldots, \ a_n - a_n a_1.
   \]

2. **Express each term in a simplified form:**
   Each term \( a_i - a_i a_{i+1} \) can be rewritten as:
   \[
   a_i (1 - a_{i+1}).
   \]
   Since \( a_i \in [0,1] \), it follows that \( a_i (1 - a_{i+1}) \geq 0 \).

3. **Consider the product of all terms:**
   We look at the product of these \( n \) expressions:
   \[
   \prod_{i=1}^n (a_i - a_i a_{i+1}) = \prod_{i=1}^n a_i (1 - a_{i+1}).
   \]

4. **Apply the AM-GM inequality:**
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we have:
   \[
   \sqrt[n]{\prod_{i=1}^n a_i (1 - a_{i+1})} \leq \frac{1}{n} \sum_{i=1}^n a_i (1 - a_{i+1}).
   \]
   Since \( a_i \in [0,1] \), the maximum value of \( a_i (1 - a_{i+1}) \) occurs when \( a_i = \frac{1}{2} \) and \( 1 - a_{i+1} = \frac{1}{2} \), giving:
   \[
   a_i (1 - a_{i+1}) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}.
   \]

5. **Calculate the maximum value:**
   If all \( a_i = \frac{1}{2} \), then:
   \[
   \prod_{i=1}^n a_i (1 - a_{i+1}) = \left( \frac{1}{4} \right)^n.
   \]
   Taking the \( n \)-th root, we get:
   \[
   \sqrt[n]{\left( \frac{1}{4} \right)^n} = \frac{1}{4}.
   \]

6. **Verify the conditions for equality:**
   The equality in the AM-GM inequality holds when all terms are equal, i.e., \( a_i = \frac{1}{2} \) for all \( i \). This indeed gives:
   \[
   a_i - a_i a_{i+1} = \frac{1}{2} \left( 1 - \frac{1}{2} \right) = \frac{1}{4}.
   \]

7. **Conclusion:**
   Therefore, the maximum value of the smallest of the numbers \( a_1 - a_1 a_2, a_2 - a_2 a_3, \ldots, a_n - a_n a_1 \) is \( \frac{1}{4} \).

The final answer is \(\boxed{\frac{1}{4}}\).