3. Given the set $S=\{1,2,3, \cdots, N\}$ and its four 500-element subsets $A_{1}, A_{2}, A_{3}, A_{4}$ that satisfy: for any $x, y \in S$, there exists some $i \in\{1,2,3,4\}$, such that $x, y \in A_{i}$. Find the maximum possible value of the positive integer $N$.
The required maximum value is 833.
On the one hand, when $N=833$, let $X=\{1,2, \cdots, 333\}, Y=\{334,335, \cdots, 500\}$, $Z=\{501,502, \cdots, 667\}, W=\{668,669, \cdots, 833\}$.
Then $|X|=333,|Y|=|Z|=167,|W|=166$, and $S=X \bigcup Y \bigcup Z \bigcup W$.
Consider four sets $A_{1}=X \bigcup Y, A_{2}=X \bigcup Z, A_{3}=X \bigcup W \bigcup\{a\}, A_{4}=Y \bigcup Z \bigcup W$, where $a \in Y \cup Z$. For any $x, y \in S$, if one of $x, y$ belongs to $X$, then $x, y$ belong to at least one of $A_{1}, A_{2}, A_{3}$; if neither $x, y$ belongs to $X$, then $x, y$ belong to $A_{4}$. This way, we obtain four 500-element subsets of $S$ that satisfy the requirements.

On the other hand, if $N \geqslant 834$, then there do not exist four 500-element subsets of $S$ that satisfy the requirements. Proof by contradiction, assume there exist four 500-element subsets $A_{1}, A_{2}, A_{3}, A_{4}$ of $S$ that satisfy the requirements. Clearly, each element of $S$ belongs to at least two of these subsets. Let $a$ be the number of elements that belong to exactly two of these subsets, then the remaining $N-a$ elements belong to at least three of these subsets.
Thus, $500 \times 4 \geqslant 2 a+3(N-a) \geqslant 3 \times 834-a=2502-a \Rightarrow a \geqslant 502$.
Let the set of these $a$ elements be $T$, then $|T|=a \geqslant 502$. Divide the elements of $T$ into 6 categories based on the subsets $A_{i}, A_{j}$ they belong to. For $1 \leqslant i<j \leqslant 4$, let $A_{ij}$ be the set of elements $x$ in $T$ that belong exactly to $A_{i}, A_{j}$.
$A_{12}$ and $A_{34}$ must have one that is an empty set (otherwise, if $x \in A_{12}, y \in A_{34}$, clearly $x \neq y$, at this time $x, y$ do not belong to any of $A_{1}, A_{2}, A_{3}, A_{4}$, contradiction), assume $A_{34}=\varnothing$. Similarly, $A_{13}$ and $A_{24}$ must have one that is an empty set, assume $A_{24}=\varnothing, A_{14}$ and $A_{23}$ must also have one that is an empty set. If $A_{23}=\varnothing$, then $T=A_{12} \bigcup A_{13} \bigcup A_{14} \subseteq A_{1} \Rightarrow\left|A_{1}\right| \geqslant|T| \geqslant 502$, contradiction; if $A_{14}=\varnothing$, then $T=A_{12} \bigcup A_{13} \cup A_{23}$, thus all elements in $A_{4}$ are those that appear at least three times in $A_{1}, A_{2}, A_{3}, A_{4}$, hence each element in $S$ appears at least twice in $A_{1}, A_{2}, A_{3}$, this way we have $|S| \leqslant \frac{1}{2}\left(\left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|\right)=750$, contradiction.