Let $m$ be a positive integer and $n=m^2+1$. Determine all real numbers $x_1,x_2,\dotsc ,x_n$ satisfying
$$x_i=1+\frac{2mx_i^2}{x_1^2+x_2^2+\cdots +x_n^2}\quad \text{for all }i=1,2,\dotsc ,n.$$
1. **Define the constant \( c \):**
   Let \( c = \frac{2m}{x_1^2 + x_2^2 + \cdots + x_n^2} \). Given that \( c > 0 \), we can rewrite the given equation as:
   \[
   x_i = 1 + c x_i^2 \quad \text{for all } i = 1, 2, \ldots, n.
   \]

2. **Form the quadratic equation:**
   Rearrange the equation to form a quadratic equation:
   \[
   c x_i^2 - x_i + 1 = 0.
   \]
   The solutions to this quadratic equation are:
   \[
   x_i = \frac{1 \pm \sqrt{1 - 4c}}{2c}.
   \]

3. **Determine the possible values of \( x_i \):**
   Since \( x_i \geq 1 \) for all \( i \), we need to consider the roots of the quadratic equation. Let \( a \) and \( b \) be the roots, where \( a \neq b \) and \( a, b > 1 \).

4. **Sum the equations:**
   Summing up all the equations for \( i = 1, 2, \ldots, n \) gives:
   \[
   \sum_{i=1}^n x_i = n + 2m.
   \]
   Let \( k_1 \) be the number of indices \( i \) with \( x_i = a \), and \( k_2 = n - k_1 \) be the number of indices \( i \) with \( x_i = b \). Without loss of generality, suppose \( k_1 \leq k_2 \).

5. **Express the sum in terms of \( a \) and \( b \):**
   \[
   k_1 a + k_2 b = n + 2m.
   \]
   Since \( n = m^2 + 1 \), we have:
   \[
   k_1 a + k_2 b = (m^2 + 1) + 2m = m^2 + 2m + 1 = (m + 1)^2.
   \]

6. **Simplify the equation:**
   \[
   k_1 (a - 1) + k_2 (b - 1) = 2m.
   \]

7. **Use the product of the roots:**
   Since \( a \) and \( b \) are roots of \( c x^2 - x + 1 = 0 \), we have:
   \[
   ab = \frac{1}{c}.
   \]
   Also, since \( a + b = \frac{1}{c} \), we have:
   \[
   (a - 1)(b - 1) = 1.
   \]

8. **Apply the AM-GM inequality:**
   Using the AM-GM inequality, we get:
   \[
   k_1 (a - 1) + k_2 (b - 1) \geq 2 \sqrt{k_1 k_2} \geq 2m.
   \]
   Equality holds if and only if \( k_1 (a - 1) = k_2 (b - 1) \) for the left inequality, and \( k_1 = 1, k_2 = m^2 \) for the right inequality.

9. **Solve for \( a \) and \( b \):**
   From \( k_1 (a - 1) = k_2 (b - 1) \) and \( k_1 = 1, k_2 = m^2 \), we get:
   \[
   a - 1 = m^2 (b - 1).
   \]
   Since \( (a - 1)(b - 1) = 1 \), we have:
   \[
   a = m + 1 \quad \text{and} \quad b = \frac{1}{m} + 1.
   \]

10. **Conclusion:**
    Thus, the solutions are:
    \[
    \boxed{\left(m + 1, \frac{1}{m} + 1, \frac{1}{m} + 1, \ldots, \frac{1}{m} + 1\right)}
    \]
    and its permutations.