Let $m,n$ be positive integers, greater than 2.Find the number of polynomials of degree $2n-1$ with distinct coefficients from the set $\left\{ 1,2,\ldots,m\right\}$ which are divisible by $x^{n-1}+x^{n-2}+\ldots+1.$
1. **Define the Polynomial and Modulo Condition:**
   Let \( P(x) = a_{2n-1}x^{2n-1} + a_{2n-2}x^{2n-2} + \cdots + a_1x + a_0 \). We need \( P(x) \) to be divisible by \( x^{n-1} + x^{n-2} + \cdots + 1 \).

2. **Simplify Using Modulo:**
   Notice that \( x^n \equiv 1 \pmod{x^{n-1} + x^{n-2} + \cdots + 1} \). Therefore,
   \[
   P(x) \equiv (a_{2n-1} + a_{n-1})x^{n-1} + (a_{2n-2} + a_{n-2})x^{n-2} + \cdots + (a_n + a_0) \pmod{x^{n-1} + \cdots + 1}.
   \]

3. **Define New Coefficients:**
   Let \( b_k = a_k + a_{n+k} \) for \( k = 0, 1, \ldots, n-1 \). We need \( R(x) = \sum_{k=0}^{n-1} b_k x^k \) to be divisible by \( x^{n-1} + \cdots + 1 \).

4. **Condition for Divisibility:**
   For \( R(x) \) to be divisible by \( x^{n-1} + \cdots + 1 \), all coefficients \( b_k \) must be equal. Let \( c = a_{2n-1} + a_{n-1} = a_{2n-2} + a_{n-2} = \cdots = a_n + a_0 \).

5. **Counting Ordered Pairs:**
   We need to count the number of ways to form ordered pairs \((x, y)\) such that \( x + y = c \) with \( 1 \leq x, y \leq m \). Let \( f(k) \) be the number of such pairs for a given \( k \).

6. **Summing Over All Possible Values:**
   The total number of polynomials is given by summing \( f(k)^n \) over all possible values of \( k \):
   \[
   \sum_{k=2}^{2m} f(k)^n.
   \]

7. **Calculate \( f(k) \):**
   For \( k \) in the range \( 2 \leq k \leq m+1 \), \( f(k) = k-1 \). For \( k \) in the range \( m+2 \leq k \leq 2m \), \( f(k) = 2m+1-k \).

8. **Final Summation:**
   Therefore, the number of polynomials is:
   \[
   \sum_{k=2}^{m+1} (k-1)^n + \sum_{k=m+2}^{2m} (2m+1-k)^n.
   \]

The final answer is \( \boxed{ \sum_{k=2}^{m+1} (k-1)^n + \sum_{k=m+2}^{2m} (2m+1-k)^n } \).