There are 2008 bags numbered from 1 to 2008, with 2008 frogs in each one of them. Two people play in turns. A play consists in selecting  a bag and taking out of it any number of frongs (at least one), leaving $ x$ frogs in it ($ x\geq 0$). After each play, from each bag with a number higher than the selected one and having more than $ x$ frogs, some frogs scape until there are $ x$ frogs in the bag. The player that takes out the last frog from bag number 1 looses. Find and explain a winning strategy.
1. **Initial Setup and First Move:**
   - There are 2008 bags, each containing 2008 frogs.
   - The first player removes 2007 frogs from bag number 2. This leaves:
     - 2008 frogs in bag number 1.
     - 1 frog in bag number 2.
     - 1 frog in each of the bags from 3 to 2008 (since frogs escape to match the number of frogs in bag number 2).

2. **Symmetric Strategy:**
   - The first player aims to maintain a symmetric strategy to ensure the second player is always forced into a losing position.
   - If the second player takes out frogs from bag number \(a+1\), the first player should adjust the number of frogs in bag number 1 to \(a\). This results in:
     - \(a\) frogs in bag number 1.
     - 1 frog in each of the bags from 2 to \(a\).
   - If the second player leaves \(a\) frogs in bag number 1, the first player should take out frogs from bag number \(a+1\). This results in:
     - \(a\) frogs in bag number 1.
     - 1 frog in each of the bags from 2 to \(a\).

3. **Ensuring Even Number of Frogs:**
   - By following the symmetric strategy, the first player ensures that an even number of frogs are taken out or escape in each turn.
   - This process continues until only 1 frog remains in bag number 1.

4. **Endgame:**
   - When only 1 frog remains in bag number 1, it will be the second player's turn to move.
   - The second player will be forced to take the last frog from bag number 1, thus losing the game.

\[
\boxed{\text{The first player has a winning strategy by following the symmetric approach.}}
\]