Let $n>k>1$ be positive integers. Determine all positive real numbers $a_1, a_2, \ldots, a_n$ which satisfy
$$
\sum_{i=1}^n \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}}=\sum_{i=1}^n a_i=n .
$$
To determine all positive real numbers \(a_1, a_2, \ldots, a_n\) that satisfy the given conditions, we start by analyzing the given equation:

\[
\sum_{i=1}^n \sqrt{\frac{k a_i^k}{(k-1) a_i^k+1}} = \sum_{i=1}^n a_i = n.
\]

1. **Applying the Arithmetic Mean-Geometric Mean (AM-GM) Inequality:**

   The AM-GM inequality states that for any non-negative real numbers \(x_1, x_2, \ldots, x_n\),

   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n},
   \]

   with equality if and only if \(x_1 = x_2 = \cdots = x_n\).

2. **Applying the Cauchy-Schwarz (CS) Inequality:**

   The Cauchy-Schwarz inequality states that for any sequences of real numbers \(a_i\) and \(b_i\),

   \[
   \left( \sum_{i=1}^n a_i b_i \right)^2 \leq \left( \sum_{i=1}^n a_i^2 \right) \left( \sum_{i=1}^n b_i^2 \right).
   \]

3. **Simplifying the given expression:**

   We start by simplifying the term inside the sum:

   \[
   \sqrt{\frac{k a_i^k}{(k-1) a_i^k + 1}}.
   \]

   Notice that:

   \[
   \frac{k a_i^k}{(k-1) a_i^k + 1} \leq \frac{k a_i^k}{k a_i^k} = \frac{k a_i^k}{k a_i^k} = 1.
   \]

   Therefore,

   \[
   \sqrt{\frac{k a_i^k}{(k-1) a_i^k + 1}} \leq \sqrt{1} = 1.
   \]

4. **Using the given condition \(\sum_{i=1}^n a_i = n\):**

   Since \(\sum_{i=1}^n a_i = n\), we can use the AM-GM inequality:

   \[
   \sum_{i=1}^n a_i = n \implies \frac{a_1 + a_2 + \cdots + a_n}{n} = 1.
   \]

   By AM-GM, we have:

   \[
   \frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}.
   \]

   Since the arithmetic mean is equal to 1, the geometric mean must also be 1, implying:

   \[
   a_1 = a_2 = \cdots = a_n = 1.
   \]

5. **Verifying the equality condition:**

   If \(a_i = 1\) for all \(i\), then:

   \[
   \sqrt{\frac{k \cdot 1^k}{(k-1) \cdot 1^k + 1}} = \sqrt{\frac{k}{k}} = 1.
   \]

   Therefore,

   \[
   \sum_{i=1}^n \sqrt{\frac{k \cdot 1^k}{(k-1) \cdot 1^k + 1}} = \sum_{i=1}^n 1 = n.
   \]

   This satisfies the given equation.

Thus, the only solution is \(a_i = 1\) for all \(i\).

The final answer is \( \boxed{ a_i = 1 } \) for all \(i\).