Example 2 Find the maximum value of the positive integer $r$ that satisfies the following condition: for any five 500-element subsets of the set $\{1,2, \cdots, 1000\}$, there exist two subsets that have at least $r$ elements in common. ${ }^{\text {[2] }}$
(2013, Romanian National Team Selection Exam)
【Analysis】Similarly, map the five subsets of 500 elements each to five vectors in a 1000-dimensional linear space. Since the requirement is the number of elements rather than their parity, we can consider the Euclidean space.
Let $v_{1}, v_{2}, v_{3}, v_{4}, v_{5}$ be the five vectors after transformation.
Notice,
$$
\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}=\sum_{i=1}^{5}\left|v_{i}\right|^{2}+2 \sum_{1<i<j \in s} v_{i} \cdot v_{j} \text {, (1) }
$$

and $\left|v_{i}\right|^{2}=500$.
Therefore, to minimize the maximum value of $v_{i} \cdot v_{j}$, we need $\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}$ to be as small as possible, and the values of $v_{i} \cdot v_{j}$ to be as evenly distributed as possible.
First, find the minimum value of $\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}$.
Let $v_{1}+v_{2}+\cdots+v_{5}=v=\left(\alpha_{1}, \alpha_{2}, \cdots, \alpha_{1000}\right)$, where $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{1000} \in \mathbf{N}$, and their sum is 2500.

Since $\left|v_{1}+v_{2}+\cdots+v_{5}\right|^{2}=\sum_{i=1}^{1000} \alpha_{i}^{2}$, the minimum value of the right-hand side is conjectured to be when 500 of $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{1000}$ are 2 and the other 500 are 3.
This can be proven using the adjustment method, or simply by
$$
\left(\alpha_{i}-2\right)\left(\alpha_{i}-3\right) \geqslant 0 \Rightarrow \alpha_{i}^{2} \geqslant 5 \alpha_{i}-6 \text {. }
$$

Summing over $i=1,2, \cdots, 1000$ gives
$$
\sum_{i=1}^{1000} \alpha_{i}^{2} \geqslant 5 \times 2500-6000=6500 \text {. }
$$

Combining equation (1) and $\left|v_{i}\right|^{2}=500$, we get
$$
\sum_{1 \leqslant i<j \leqslant 5} v_{i} \cdot v_{j} \geqslant 2000 \Rightarrow \max _{1 \leqslant i<j \leqslant 5} v_{i} \cdot v_{j} \geqslant 200 \text {. }
$$

Next, consider the construction part. We only need to construct five vectors in a 10-dimensional space, each with five components being 1 and the other five components being 0, and the inner product of any two vectors is 2. Then, by replicating each of these five vectors 100 times, we obtain the required vectors in the 1000-dimensional space.

Using the cyclic construction method (the first five components cycle, and the last five components cycle), it is not difficult to construct the five vectors as:
$$
\begin{array}{l}
\boldsymbol{w}_{1}=(1,1,1,0,0,1,0,1,0,0), \\
w_{2}=(0,1,1,1,0,0,1,0,1,0), \\
w_{3}=(0,0,1,1,1,0,0,1,0,1), \\
w_{4}=(1,0,0,1,1,1,0,0,1,0), \\
w_{5}=(1,1,0,0,1,0,1,0,0,1) .
\end{array}
$$

In summary, the maximum value of $r$ is 200.