Let $n$ be a positive integer with $n \geq 2$, and $0<a_{1}, a_{2},...,a_{n}< 1$. Find the maximum value of the sum
$\sum_{i=1}^{n}(a_{i}(1-a_{i+1}))^{\frac{1}{6}}$
where $a_{n+1}=a_{1}$
1. We start with the given sum:
   \[
   \sum_{i=1}^{n} \left( a_i (1 - a_{i+1}) \right)^{\frac{1}{6}}
   \]
   where \( a_{n+1} = a_1 \).

2. We can rewrite the term inside the sum using the properties of exponents:
   \[
   \left( a_i (1 - a_{i+1}) \right)^{\frac{1}{6}} = \left( 2^{-\frac{3}{2}} \cdot 2^{\frac{3}{2}} a_i (1 - a_{i+1}) \right)^{\frac{1}{6}}
   \]
   Simplifying the expression inside the parentheses:
   \[
   = 2^{-\frac{3}{2} \cdot \frac{1}{6}} \cdot \left( 2^{\frac{3}{2}} a_i (1 - a_{i+1}) \right)^{\frac{1}{6}}
   \]
   \[
   = 2^{-\frac{1}{4}} \cdot \left( 2^{\frac{3}{2}} a_i (1 - a_{i+1}) \right)^{\frac{1}{6}}
   \]

3. Using the AM-GM inequality, we know that for any non-negative numbers \( x_1, x_2, \ldots, x_k \):
   \[
   \frac{x_1 + x_2 + \cdots + x_k}{k} \geq \sqrt[k]{x_1 x_2 \cdots x_k}
   \]
   Applying this to our expression, we get:
   \[
   \left( 2^{\frac{3}{2}} a_i (1 - a_{i+1}) \right)^{\frac{1}{6}} \leq \frac{2^{\frac{3}{2}} a_i + 2^{\frac{3}{2}} (1 - a_{i+1})}{6}
   \]

4. Summing over all \( i \) from 1 to \( n \):
   \[
   \sum_{i=1}^{n} \left( a_i (1 - a_{i+1}) \right)^{\frac{1}{6}} \leq \sum_{i=1}^{n} \frac{2^{\frac{3}{2}} a_i + 2^{\frac{3}{2}} (1 - a_{i+1})}{6}
   \]

5. Simplifying the right-hand side:
   \[
   = \frac{2^{\frac{3}{2}}}{6} \sum_{i=1}^{n} (a_i + 1 - a_{i+1})
   \]
   Since \( a_{n+1} = a_1 \), the terms \( a_i \) and \( -a_{i+1} \) will cancel out, leaving us with:
   \[
   = \frac{2^{\frac{3}{2}}}{6} \sum_{i=1}^{n} 1
   \]
   \[
   = \frac{2^{\frac{3}{2}}}{6} \cdot n
   \]
   \[
   = \frac{2^{\frac{3}{2}} n}{6}
   \]
   \[
   = \frac{2 \sqrt{2} n}{6}
   \]
   \[
   = \frac{\sqrt{2} n}{3}
   \]

6. Therefore, the maximum value of the sum is:
   \[
   \boxed{\frac{\sqrt{2} n}{3}}
   \]