Find all reals $\lambda$ for which there is a nonzero polynomial $P$ with real coefficients such that
$$\frac{P(1)+P(3)+P(5)+\ldots+P(2n-1)}n=\lambda P(n)$$for all positive integers $n$, and find all such polynomials for $\lambda=2$.
1. **Given Equation and Initial Analysis:**
   We start with the given equation:
   \[
   \frac{P(1) + P(3) + P(5) + \ldots + P(2n-1)}{n} = \lambda P(n)
   \]
   for all positive integers \( n \). We need to find all real values of \( \lambda \) for which there exists a nonzero polynomial \( P \) with real coefficients satisfying this equation.

2. **Rewriting the Equation:**
   Multiply both sides by \( n \):
   \[
   P(1) + P(3) + P(5) + \ldots + P(2n-1) = \lambda n P(n)
   \]

3. **Considering the Polynomial Degree:**
   Let \( \deg(P) = k \). We assume \( P(x) = a_k x^k + a_{k-1} x^{k-1} + \ldots + a_0 \).

4. **Analyzing the Polynomial \( P \):**
   We need to find a relationship between \( P(n) \) and \( P(2n+1) \). Consider the difference:
   \[
   \lambda (n+1) P(n+1) - \lambda n P(n) = P(2n+1)
   \]
   This must hold for all positive integers \( n \).

5. **Degree Consideration:**
   Since \( P \) is a polynomial of degree \( k \), the degree of \( P(n+1) \) and \( P(n) \) is \( k \), and the degree of \( P(2n+1) \) is also \( k \). Therefore, the degrees on both sides of the equation must match.

6. **Finding \( \lambda \):**
   By comparing the leading coefficients, we find:
   \[
   \lambda (n+1)^k - \lambda n^k = (2n+1)^k
   \]
   For large \( n \), the dominant term on the left is \( \lambda k n^{k-1} \) and on the right is \( 2^k n^k \). Equating these gives:
   \[
   \lambda k n^{k-1} = 2^k n^k
   \]
   Solving for \( \lambda \):
   \[
   \lambda = \frac{2^k}{k+1}
   \]

7. **Existence of Polynomial \( P \):**
   The solution space is a vector space. If \( \lambda \) is of the form \( \frac{2^k}{k+1} \), the linear map \( P(x) \to \lambda (x+1)P(x+1) - \lambda xP(x) - P(2x+1) \) defines a map from the space of polynomials of degree at most \( k \) to the space of polynomials of degree at most \( k-1 \). This map has a non-trivial kernel, so such a polynomial \( P \) does exist.

8. **Case for \( \lambda = 2 \):**
   For \( \lambda = 2 \):
   \[
   2 = \frac{2^k}{k+1}
   \]
   Solving for \( k \):
   \[
   2(k+1) = 2^k \implies k+1 = 2^{k-1}
   \]
   Testing values, we find \( k = 3 \) satisfies this equation.

9. **Form of Polynomial \( P \):**
   For \( k = 3 \), \( P \) is a cubic polynomial. We have:
   \[
   P(x) = a(x^2 - 1)(x + b)
   \]
   Setting \( x = 0 \) gives \( P(-1) = 0 \), and setting \( x = 1 \) gives \( P(1) = 0 \). Therefore, \( P(x) = a(x^2 - 1)x \).

10. **Verification:**
    Substituting \( P(x) = a(x^2 - 1)x \) into the original equation confirms that it satisfies the given condition for \( \lambda = 2 \).

The final answer is \( \boxed{ \lambda = \frac{2^k}{k+1} } \) and for \( \lambda = 2 \), \( P(x) = a(x^2 - 1)x \).