Two players alternately replace the stars in the expression 
	\[*x^{2000}+*x^{1999}+...+*x+1 \]
by real numbers. The player who makes the last move loses if the resulting polynomial has a real root $t$ with $|t| < 1$, and wins otherwise. Give a winning strategy for one of the players.
1. **Identify the players and their roles**: Let \( A \) be the first player and \( B \) be the second player. The players alternately replace the stars in the polynomial \( *x^{2000} + *x^{1999} + \ldots + *x + 1 \) with real numbers. The player who makes the last move loses if the resulting polynomial has a real root \( t \) with \( |t| < 1 \), and wins otherwise.

2. **Determine the winning strategy based on the parity of \( n \)**:
   - If \( n = 2k \) (even), \( A \) has the winning strategy.
   - If \( n = 2k + 1 \) (odd), \( B \) has the winning strategy.

3. **Analyze the case when \( n = 2k \)**:
   - Denote the asterisk before \( x^k \) as the \( k \)-index asterisk.
   - When \( n = 2k \), the number of odd-index asterisks equals the number of even-index asterisks.
   - Player \( A \) can always perform the replacement such that after each replacement, the number of even-index asterisks does not exceed the number of odd-index asterisks.
   - After \( 2k - 2 \) replacements, there are only two asterisks left, let them be the \( t \)-index asterisk and the \( p \)-index asterisk, where \( p \) is odd.

4. **Prove that \( A \) can force a win**:
   - We need to show that there exists \( a_t \) such that the polynomial \( P(x) \) has a real root \( \alpha \) with \( |\alpha| < 1 \) for all \( a_i \) where \( i \ne t \).
   - Set \( a, b > 0 \) such that \( a \ne b \) and \( [-a, b] \subset (-1, 1) \).
   - Let \( \gamma = \left( \frac{b}{a} \right)^p > 0 \).
   - Set \( a_t = \frac{-\gamma Q(-a) - Q(b)}{\gamma (-a)^t + b^t} \), where \( Q(x) = \sum_{i \ne t, p} a_i x^i \).
   - Since \( a \ne b \) and \( t \ne p \), \( \gamma (-a)^t + b^t \ne 0 \).

5. **Verify the polynomial has a root in \((-1, 1)\)**:
   - Consider the polynomial \( P(x) \):
     \[
     P(x) = Q(x) + a_t x^t + a_p x^p
     \]
   - Evaluate \( P(x) \) at \( -a \) and \( b \):
     \[
     \gamma P(-a) + P(b) = \gamma [Q(-a) + a_t (-a)^t + a_p (-a)^p] + [Q(b) + a_t b^t + a_p b^p]
     \]
   - Substitute \( a_t \):
     \[
     \gamma P(-a) + P(b) = \gamma Q(-a) + Q(b) + a_t [\gamma (-a)^t + b^t] = 0
     \]
   - Hence, \( P(-a) P(b) \leq 0 \) (since \( \gamma > 0 \)).
   - Therefore, \( P(x) \) always has a real root in \( (-1, 1) \).

6. **Conclusion for \( n = 2k \)**:
   - Player \( A \) can always ensure that the polynomial has a real root in \( (-1, 1) \) after the last move, thus winning the game.

7. **Analyze the case when \( n = 2k + 1 \)**:
   - Player \( B \) can always perform the replacement such that the number of even-index asterisks does not exceed the number of odd-index asterisks.
   - By a similar argument as above, \( B \) can ensure that the polynomial has a real root in \( (-1, 1) \) after the last move, thus winning the game.

The final answer is \( \boxed{ A } \) wins if \( n = 2k \) and \( B \) wins if \( n = 2k + 1 \).