Let $n$ and $k$ be two integers which are greater than $1$. Let $a_1,a_2,\ldots,a_n,c_1,c_2,\ldots,c_m$ be non-negative real numbers such that
i) $a_1\ge a_2\ge\ldots\ge a_n$ and $a_1+a_2+\ldots+a_n=1$;
ii) For any integer $m\in\{1,2,\ldots,n\}$, we have that $c_1+c_2+\ldots+c_m\le m^k$.
Find the maximum of $c_1a_1^k+c_2a_2^k+\ldots+c_na_n^k$.
To find the maximum of \( c_1a_1^k + c_2a_2^k + \ldots + c_na_n^k \), we will use the given conditions and properties of the sequences \( \{a_i\} \) and \( \{c_i\} \).

1. **Given Conditions:**
   - \( a_1 \ge a_2 \ge \ldots \ge a_n \) and \( a_1 + a_2 + \ldots + a_n = 1 \).
   - For any integer \( m \in \{1, 2, \ldots, n\} \), \( c_1 + c_2 + \ldots + c_m \le m^k \).

2. **Expression to Maximize:**
   \[
   \sum_{i=1}^n c_i a_i^k
   \]

3. **Upper Bound Analysis:**
   We start by considering the sum \( \sum_{i=1}^n c_i a_i^k \). Using the given conditions, we can derive an upper bound for this sum.

   \[
   \sum_{i=1}^n c_i a_i^k = a_n^k \sum_{i=1}^n c_i + \sum_{i=1}^{n-1} \left( \sum_{j=1}^i c_j \right) (a_i^k - a_{i+1}^k)
   \]

   Given \( \sum_{i=1}^m c_i \le m^k \), we can write:

   \[
   \sum_{i=1}^n c_i \le n^k
   \]

   Therefore,

   \[
   a_n^k \sum_{i=1}^n c_i \le a_n^k n^k
   \]

   For the remaining terms, we use the fact that \( \sum_{j=1}^i c_j \le i^k \):

   \[
   \sum_{i=1}^{n-1} \left( \sum_{j=1}^i c_j \right) (a_i^k - a_{i+1}^k) \le \sum_{i=1}^{n-1} i^k (a_i^k - a_{i+1}^k)
   \]

4. **Summation Simplification:**
   Notice that the sum \( \sum_{i=1}^{n-1} i^k (a_i^k - a_{i+1}^k) \) can be telescoped:

   \[
   \sum_{i=1}^{n-1} i^k (a_i^k - a_{i+1}^k) = \sum_{i=1}^{n-1} i^k a_i^k - \sum_{i=2}^n (i-1)^k a_i^k
   \]

   This simplifies to:

   \[
   \sum_{i=1}^{n-1} i^k a_i^k - \sum_{i=2}^n (i-1)^k a_i^k = \sum_{i=1}^n i^k a_i^k - \sum_{i=1}^{n-1} (i+1)^k a_{i+1}^k
   \]

   Combining these terms, we get:

   \[
   \sum_{i=1}^n i^k a_i^k - \sum_{i=2}^n (i-1)^k a_i^k = a_1^k + \sum_{i=2}^n (i^k - (i-1)^k) a_i^k
   \]

5. **Final Expression:**
   Since \( a_1 + a_2 + \ldots + a_n = 1 \), we can use the fact that \( a_1^k + \sum_{i=2}^n (i^k - (i-1)^k) a_i^k \le 1 \):

   \[
   \sum_{i=1}^n c_i a_i^k \le 1
   \]

6. **Equality Condition:**
   Equality holds when \( \sum_{i=1}^s c_i = s^k \) and \( a_1 = a_2 = \ldots = a_s = \frac{1}{s} \), with \( a_{s+1} = \ldots = a_n = 0 \), for some \( s \in \mathbb{Z} \) and \( 1 \le s \le n \).

The final answer is \( \boxed{ 1 } \)