Given are the sequences
\[ (..., a_{-2}, a_{-1}, a_0, a_1, a_2, ...);  (..., b_{-2}, b_{-1}, b_0, b_1, b_2, ...); (..., c_{-2}, c_{-1}, c_0, c_1, c_2, ...)\]
of positive real numbers. For each integer $n$ the following inequalities hold:
\[a_n \geq \frac{1}{2} (b_{n+1} + c_{n-1})\]
\[b_n \geq \frac{1}{2} (c_{n+1} + a_{n-1})\]
\[c_n \geq \frac{1}{2} (a_{n+1} + b_{n-1})\]
Determine $a_{2005}$, $b_{2005}$, $c_{2005}$, if $a_0 = 26, b_0 = 6, c_0 = 2004$.
1. **Define the sequences cyclically:**
   We set \( x_n \) to be \( a_n \) if \( n \equiv 0 \pmod{3} \), \( b_n \) if \( n \equiv 1 \pmod{3} \), and \( c_n \) if \( n \equiv 2 \pmod{3} \). Similarly, we define \( y_n \) and \( z_n \) by cyclically permuting \( a, b, c \).

2. **Establish the inequalities for \( x_n \):**
   From the given inequalities, we have:
   \[
   a_n \geq \frac{1}{2} (b_{n+1} + c_{n-1})
   \]
   \[
   b_n \geq \frac{1}{2} (c_{n+1} + a_{n-1})
   \]
   \[
   c_n \geq \frac{1}{2} (a_{n+1} + b_{n-1})
   \]
   By substituting the cyclic definitions, we get:
   \[
   x_n \geq \frac{1}{2} (x_{n-1} + x_{n+1})
   \]

3. **Analyze the inequality \( x_n \geq \frac{1}{2} (x_{n-1} + x_{n+1}) \):**
   This inequality implies:
   \[
   2x_n \geq x_{n-1} + x_{n+1}
   \]
   Rearranging, we get:
   \[
   x_n - x_{n-1} \geq x_{n+1} - x_n
   \]
   This means that the sequence \( x_n \) is concave.

4. **Prove that \( x_n \) is constant:**
   Suppose \( x_{n+1} - x_n > 0 \). Then, as \( n \to -\infty \), \( x_n \) would increase without bound, which contradicts the boundedness of the sequence. Similarly, if \( x_{n+1} - x_n < 0 \), then as \( n \to \infty \), \( x_n \) would decrease without bound, which is also a contradiction. Therefore, \( x_{n+1} - x_n = 0 \), implying that \( x_n \) is constant.

5. **Apply the same reasoning to \( y_n \) and \( z_n \):**
   By similar arguments, we can show that \( y_n \) and \( z_n \) are also constant sequences.

6. **Determine the constants:**
   Given the initial conditions \( a_0 = 26 \), \( b_0 = 6 \), \( c_0 = 2004 \), and the cyclic nature of the sequences, we have:
   \[
   (a_n, b_n, c_n) = (b_{n+1}, c_{n+1}, a_{n+1}) = (c_{n+2}, a_{n+2}, b_{n+2}) = (a_{n+3}, b_{n+3}, c_{n+3})
   \]
   Therefore, the sequences repeat every 3 steps.

7. **Find \( a_{2005}, b_{2005}, c_{2005} \):**
   Since \( 2005 \equiv 2 \pmod{3} \), we have:
   \[
   a_{2005} = c_0 = 2004
   \]
   \[
   b_{2005} = a_0 = 26
   \]
   \[
   c_{2005} = b_0 = 6
   \]

The final answer is \( (a_{2005}, b_{2005}, c_{2005}) = \boxed{(2004, 26, 6)} \).