A student is playing computer. Computer shows randomly 2002 positive numbers. Game's rules let do the following operations 
- to take 2 numbers from these, to double first one, to add the second one and to save the sum. 
- to take another 2 numbers from the remainder numbers, to double the first one, to add the second one, to multiply this sum with previous and to save the result. 
- to repeat this procedure, until all the 2002 numbers won't be used. 
Student wins the game if final product is maximum possible. 
Find the winning strategy and prove it.
1. **Initial Setup and Assumptions**:
   - We are given 2002 positive numbers.
   - We need to maximize the final product by performing a series of operations.
   - Without loss of generality (WLOG), assume the numbers are sorted in non-increasing order: \( a_1 \geq a_2 \geq \ldots \geq a_{2002} \).

2. **Claim 1: The numbers that are doubled are precisely the largest 1001 numbers in \(\{a_n\}\)**:
   - **Proof by Contradiction**:
     - Suppose the maximal product \( P \) has a doubled number \( a_i \) not in \(\{a_1, a_2, \ldots, a_{1001}\}\).
     - In \( P \), there exist two factors \((2a_i + a_j)\) and \((2a_k + a_l)\).
     - For maximality, \( a_j \) must be less than or equal to \( a_i \); otherwise, switching the two numbers would yield a larger product.
     - By the pigeonhole principle, there exist numbers \( a_k, a_l \) where \( a_k, a_l \in \{a_1, a_2, \ldots, a_{1001}\} \).
     - So, \( P = (2a_i + a_j)(2a_k + a_l)K = (4a_i a_k + 2a_i a_l + 2a_j a_k + a_j a_l)K \).
     - Define \( Q = (2a_l + a_j)(2a_k + a_i)K = (4a_k a_l + 2a_i a_l + 2a_j a_k + a_i a_j)K \).
     - It is clear that \( Q > P \), and \( P \) is not the largest product; contradiction.
     - Thus, the largest product cannot have any doubled numbers that are not from the set \(\{a_1, a_2, \ldots, a_{1001}\}\), and the claim is proven.

3. **Claim 2: For the maximal product \( P = (2a_1 + b_1)(2a_2 + b_2) \ldots (2a_{1001} + b_{1001}) \), we have \( b_1 \leq b_2 \leq \ldots \leq b_{1001} \)**:
   - **Proof by Contradiction**:
     - Suppose there exist two indices \( i, j \) such that \( a_i > a_j \) and \( b_i > b_j \).
     - Rewrite \( P \) as \( (2a_i + b_i)(2a_j + b_j)K = (4a_i a_j + 2a_i b_j + 2a_j b_i + b_i b_j)K \).
     - Define \( Q = (2a_i + b_j)(2a_j + b_i)K = (4a_i a_j + 2a_i b_i + 2a_j b_j + b_i b_j)K \).
     - Clearly, similarly sorted sequences imply that \( Q > P \), so there is once again a contradiction.
     - Therefore, the maximal product must not have \( b_i > b_j \) and \( i > j \) for any pairs \((i, j)\).
     - This forces \( b_1 \leq b_2 \leq \ldots \leq b_{1001} \).

4. **Conclusion**:
   - The largest possible product is achieved when the sequence \( a_1, a_2, \ldots, a_{1001} \) is paired with the sequence \( a_{2002}, a_{2001}, \ldots, a_{1002} \) respectively.
   - Therefore, the largest possible answer is \( P = (2a_1 + a_{2002})(2a_2 + a_{2001}) \ldots (2a_{1001} + a_{1002}) \).

The final answer is \( \boxed{ P = (2a_1 + a_{2002})(2a_2 + a_{2001}) \ldots (2a_{1001} + a_{1002}) } \).