492 Select five subsets $A_{1}, A_{2}, \cdots, A_{5}$ from the set $\{1,2, \cdots, 1000\}$ such that $\left|A_{i}\right|=500(i=1,2$, $\cdots, 5)$. Find the maximum value of the number of common elements in any three of these subsets.
Let $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{5}$ be five 1000-dimensional vectors, and let $\alpha_{i}(j)$ denote the $j$-th element of $\alpha_{i}$. Then $\alpha_{1}, \alpha_{2}, \cdots, \alpha_{5}$ satisfy
$$
\alpha_{i}(j)=\left\{\begin{array}{l}
1, j \in A_{i} ; \\
0, j \notin A_{i} .
\end{array}\right.
$$

Let the maximum number of common elements in any three subsets be $r$.
For any three vectors
$$
\begin{array}{l}
\boldsymbol{\beta}_{1}=\left(b_{1}, b_{2}, \cdots, b_{t}\right), \boldsymbol{\gamma}_{1}=\left(c_{1}, c_{2}, \cdots, c_{t}\right), \\
\boldsymbol{\delta}_{1}=\left(d_{1}, d_{2}, \cdots, d_{t}\right),
\end{array}
$$

define the ternary inner product operation
$$
\left(\boldsymbol{\beta}_{1}, \boldsymbol{\gamma}_{1}, \boldsymbol{\delta}_{1}\right)=b_{1} c_{1} d_{1}+b_{2} c_{2} d_{2}+\cdots+b_{t} c_{t} d_{t} .
$$

Clearly, this operation is linear and satisfies the commutative law.
By the problem, $\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}\right)=500$.
For $1 \leqslant i<j<k \leqslant 5,\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}, \boldsymbol{\alpha}_{k}\right)$ equals the number of common elements in the sets $A_{i}, A_{j}, A_{k}$.
Let $\boldsymbol{\alpha}=\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}+\boldsymbol{\alpha}_{3}+\boldsymbol{\alpha}_{4}+\boldsymbol{\alpha}_{5}$. Then
$(\boldsymbol{\alpha}, \boldsymbol{\alpha}, \boldsymbol{\alpha})$
$$
\begin{aligned}
= & \sum_{i=1}^{5}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}\right)+3 \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)+ \\
& 6 \sum_{i<j<k}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}, \boldsymbol{\alpha}_{k}\right) \\
= & 2500+3 \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)+6 \sum_{i<j<k}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}, \boldsymbol{\alpha}_{k}\right) \\
\leqslant & 2500+3 \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)+60 r .
\end{aligned}
$$

Also, $\boldsymbol{\alpha}$ is a 1000-dimensional vector, and
$$
\boldsymbol{\alpha}=\left(a_{1}, a_{2}, \cdots, a_{1000}\right)\left(a_{i} \in\{0,1, \cdots, 5\}\right),
$$

satisfying $\sum_{i=1}^{1000} a_{i}=2500$.
At this point, $(\boldsymbol{\alpha}, \boldsymbol{\alpha}, \boldsymbol{\alpha})=\sum_{i=1}^{1000} a_{i}^{3}$, and $\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)$ represents the number of common elements in the sets $A_{i}$ and $A_{j}$, so $\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)=\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)$, where $(\boldsymbol{a}, \boldsymbol{b})$ denotes the usual inner product of vectors $\boldsymbol{a}$ and $\boldsymbol{b}$.
$$
\begin{array}{l}
\text { Hence } \sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right)=\sum_{i \neq j}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{j}\right) \\
=(\boldsymbol{\alpha}, \boldsymbol{\alpha})-\sum_{i=1}^{5}\left(\boldsymbol{\alpha}_{i}, \boldsymbol{\alpha}_{i}\right) \\
=\sum_{i=1}^{1000} a_{i}^{2}-2500 .
\end{array}
$$

At this point,
$$
\begin{array}{l}
60 r \geqslant \sum_{i=1}^{1000} a_{i}^{3}-3\left(\sum_{i=1}^{1000} a_{i}^{2}-2500\right)-2500 \\
=\sum_{i=1}^{1000}\left(a_{i}^{3}-3 a_{i}^{2}\right)+5000 . \\
\text { Also, }\left(a_{i}-2\right)\left(a_{i}-3\right)\left(a_{i}+2\right) \geqslant 0 \\
\Leftrightarrow a_{i}^{3}-3 a_{i}^{2}-4 a_{i}+12 \geqslant 0 .
\end{array}
$$

Thus, for any $1 \leqslant i \leqslant 1000$,
$$
a_{i}^{3}-3 a_{i}^{2} \geqslant 4 a_{i}-12 \text {. }
$$

Therefore, $60 r \geqslant \sum_{i=1}^{1000}\left(4 a_{i}-12\right)+5000$
$$
=4 \times 2500-12000+5000=3000 \text {. }
$$

Thus, $r \geqslant 50$.
Below is a construction for $r=50$.
Divide the 1000 elements into 20 sets, each containing 50 elements. These 20 sets are divided into two groups, denoted as
$$
\begin{array}{l} 
B_{123}, B_{124}, B_{125}, B_{134}, B_{135}, B_{145}, B_{234}, B_{235}, \\
B_{245}, B_{345} ; \\
C_{12}, C_{13}, C_{14}, C_{15}, C_{23}, C_{24}, C_{25}, C_{34}, C_{35}, C_{45} .
\end{array}
$$

Define $A_{i}=\{$ the union of all sets whose indices contain $i\}$.
At this point, for any $1 \leqslant i<j<k \leqslant 5$,
$$
A_{i} \cap A_{j} \cap A_{k}=B_{i j k},
$$

which has exactly 50 elements.
(Chen Kai, Xueersi Peiyou Beijing Branch, 100086)