Let $X_1,X_2,..$ be independent random variables with the same distribution, and let $S_n=X_1+X_2+...+X_n, n=1,2,...$. For what real numbers $c$ is the following statement true:
$$P\left(\left| \frac{S_{2n}}{2n}- c \right| \leqslant \left| \frac{S_n}{n}-c\right| \right)\geqslant \frac{1}{2}$$
1. **Define the problem in terms of new variables:**
   Let \( X := \frac{X_1 + X_2 + \dots + X_n}{n} \) and \( Y := \frac{X_{n+1} + \dots + X_{2n}}{n} \). We need to prove:
   \[
   P\left(\left| \frac{X + Y}{2} - c \right| \leq \left| X - c \right| \right) \geq \frac{1}{2}
   \]
   for any real number \( c \).

2. **Simplify the problem by setting \( c = 0 \):**
   Without loss of generality, we can assume \( c = 0 \). If we prove the statement for \( c = 0 \), it will hold for any \( c \) by considering \( X' := X - c \) and \( Y' := Y - c \). Thus, we need to show:
   \[
   P\left(\left| \frac{X + Y}{2} \right| \leq \left| X \right| \right) \geq \frac{1}{2}
   \]

3. **Analyze the condition \( \left| \frac{X + Y}{2} \right| \leq \left| X \right| \):**
   The inequality \( \left| \frac{X + Y}{2} \right| \leq \left| X \right| \) can be rewritten as:
   \[
   -|X| \leq \frac{X + Y}{2} \leq |X|
   \]
   Multiplying through by 2, we get:
   \[
   -2|X| \leq X + Y \leq 2|X|
   \]
   Rearranging, we find:
   \[
   -3|X| \leq Y \leq |X|
   \]

4. **Consider the cases for \( X \leq 0 \) and \( X > 0 \):**
   - If \( X \leq 0 \), then \( |X| = -X \), and the condition becomes:
     \[
     X \leq Y \leq -3X
     \]
   - If \( X > 0 \), then \( |X| = X \), and the condition becomes:
     \[
     -3X \leq Y \leq X
     \]

5. **Express the probability \( p \) in terms of integrals:**
   Let \( f(x) \) be the probability density function of \( X \) (and \( Y \)). The probability \( p \) can be written as:
   \[
   p = \int_{-\infty}^0 f(x) \int_{x}^{-3x} f(y) \, dy \, dx + \int_{0}^{\infty} f(x) \int_{-3x}^{x} f(y) \, dy \, dx
   \]

6. **Define the region \( D \) and its subset \( D' \):**
   Let \( D \) be the region defined by:
   \[
   D := \{(x, y) \mid x \leq 0, x \leq y \leq -3x \text{ or } x > 0, -3x \leq y \leq x\}
   \]
   Let \( D' \) be the subset of \( D \) defined by:
   \[
   D' := \{(x, y) \mid x \leq 0, x \leq y \leq -x \text{ or } x > 0, -x \leq y \leq x\}
   \]

7. **Show that \( D' \subset D \) and calculate the integral over \( D' \):**
   It is clear that \( D' \subset D \). The integral over \( D' \) is:
   \[
   \iint_{D'} f(x) f(y) \, dx \, dy
   \]
   By symmetry, the integral over \( D' \) and its complement in \( \mathbb{R}^2 \) are equal:
   \[
   \iint_{D'} f(x) f(y) \, dx \, dy = \frac{1}{2}
   \]

8. **Conclude that \( p \geq \frac{1}{2} \):**
   Since \( D' \subset D \), we have:
   \[
   p = \iint_{D} f(x) f(y) \, dx \, dy \geq \iint_{D'} f(x) f(y) \, dx \, dy = \frac{1}{2}
   \]

The final answer is \( \boxed{ c \in \mathbb{R} } \)