Find all positive integers   $x , y$ which are roots of the equation 

                                                  $2 x^y-y= 2005$
[u] Babis[/u]
To solve the equation \(2x^y - y = 2005\) for positive integers \(x\) and \(y\), we can proceed as follows:

1. **Initial Observations**:
   - Since \(2x^y - y = 2005\), it is clear that \(x\) must be at least 2 because if \(x = 1\), then \(2 \cdot 1^y - y = 2 - y\), which cannot equal 2005 for any positive integer \(y\).
   - We also observe that \(y\) must be odd. This is because if \(y\) were even, say \(y = 2k\), then \(2x^{2k} - 2k = 2005\), which would imply \(2x^{2k}\) is even and \(2005 + 2k\) is odd, leading to a contradiction.

2. **Substitution**:
   - Let \(y = 2a - 1\) where \(a\) is a positive integer. Then the equation becomes:
     \[
     2x^{2a-1} - (2a - 1) = 2005
     \]
     Simplifying, we get:
     \[
     2x^{2a-1} = 2005 + 2a - 1
     \]
     \[
     2x^{2a-1} = 2004 + 2a
     \]
     \[
     x^{2a-1} = 1002 + a
     \]

3. **Inductive Argument**:
   - Assume for some \(a = a_0\), we have \(2^{2a_0-1} > 1002 + a_0\). Then for \(a = a_0 + 1\), we have:
     \[
     2^{2a_0+1} = 4 \cdot 2^{2a_0-1} > 4(1002 + a_0) > 1003 + a_0 = 1002 + (a_0 + 1)
     \]
   - Since \(x^{2a-1} \geq 2^{2a-1}\), inductively, we conclude that for all \(a \geq a_0\), the equality \(x^{2a-1} = 1002 + a\) cannot be satisfied.

4. **Finding \(a_0\)**:
   - We need to find the smallest \(a_0\) such that \(2^{2a_0-1} > 1002 + a_0\). Testing values:
     - For \(a = 6\):
       \[
       2^{11} = 2048 > 1002 + 6 = 1008
       \]
     - Therefore, \(a_0 = 6\) works, and we only need to check \(a = 1, 2, 3, 4, 5\).

5. **Checking Small Values of \(a\)**:
   - For \(a = 1\):
     \[
     x^{2 \cdot 1 - 1} = x^1 = 1002 + 1 = 1003
     \]
     \[
     x = 1003, \quad y = 1
     \]
   - For \(a = 2\):
     \[
     x^{2 \cdot 2 - 1} = x^3 = 1002 + 2 = 1004
     \]
     - \(x^3 = 1004\) does not yield an integer solution for \(x\).
   - For \(a = 3\):
     \[
     x^{2 \cdot 3 - 1} = x^5 = 1002 + 3 = 1005
     \]
     - \(x^5 = 1005\) does not yield an integer solution for \(x\).
   - For \(a = 4\):
     \[
     x^{2 \cdot 4 - 1} = x^7 = 1002 + 4 = 1006
     \]
     - \(x^7 = 1006\) does not yield an integer solution for \(x\).
   - For \(a = 5\):
     \[
     x^{2 \cdot 5 - 1} = x^9 = 1002 + 5 = 1007
     \]
     - \(x^9 = 1007\) does not yield an integer solution for \(x\).

Thus, the only solution is when \(a = 1\), giving \(x = 1003\) and \(y = 1\).

The final answer is \(\boxed{(x, y) = (1003, 1)}\).