Example 6 (30th Russian Mathematical Olympiad) On the table, there are 2004 small boxes, each containing 1 ball. It is known that some of the balls are white, and there is an even number of white balls. You are allowed to point to any 2 boxes and ask: "Do they contain at least 1 white ball?" How many times do you need to ask, at a minimum, to determine 2 boxes that both contain white balls?

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To solve the problem, we will number the boxes (and the balls inside them) from 1 to 2004, and refer to the boxes by their numbers. We will call non-white balls black balls.

First, we prove that 4005 questions can find 2 white balls. We ask the questions $(1,2),(1,3), \cdots,(1,2004)$, $(2,3),(2,4), \cdots,(2,2004)$. If all the answers are affirmative, then balls 1 and 2 are white (in fact, if ball 1 is black, then we can find at least one more black ball, which, together with ball 1, forms a black pair). If at least one answer is negative (let's assume it is a question involving ball 1), then ball 1 is black. In this case, the other ball in the questions containing ball 1 that gave an affirmative answer must be white. This way, we have identified all the white balls, and there are at least 2.

Second, we prove that fewer than 4005 questions cannot ensure finding 2 white balls. Suppose there exists a scheme that can ensure finding 2 white balls with fewer questions. We can give affirmative answers to all the questions. At this point, after at most 4004 questions, we must point out 2 boxes (for definiteness, let's say the 1st and 2nd boxes) and assert that they contain white balls. At this time, there must be some questions of the form $(1, n)$ or $(2, n)$ that have not been asked, for example, the question $(1, k)$, otherwise, 4005 questions would be needed. If we place black balls in the 1st and $k$-th boxes and white balls in all other boxes, then all the answers we gave are correct. However, the first ball pointed out by the player is black, which is a contradiction.
Therefore, at least 4005 questions are needed.