Four, (50 points) Find the maximum value of a positive integer $n$, such that for any simple graph of order $n$ with vertices $V_{1}, V_{2}, \cdots, V_{n}$, there always exist $n$ subsets $A_{1}, A_{2}, \cdots, A_{n}$ of the set $\{1,2, \cdots, 2014\}$, satisfying: $A_{i} \cap A_{j} \neq \varnothing (i \neq j)$ if and only if $V_{i}$ is adjacent to $V_{j}$.
First, prove that $n \leqslant 89$.
Suppose $n \geqslant 90$, consider the complete bipartite graph
$$
G\left(V_{1}, V_{2}, \cdots, V_{45} ; V_{1}^{\prime}, V_{2}^{\prime}, \cdots, V_{n-45}^{\prime}\right)
$$
(i.e., where $V_{i} V_{j}^{\prime}$ are all the edges), and assume that the $n$ subsets $A_{1}, A_{2}, \cdots, A_{45} ; A_{1}^{\prime}, A_{2}^{\prime}, \cdots, A_{n-45}^{\prime}$ satisfy the conditions.
Since $A_{i} \cap A_{j}^{\prime} \neq \varnothing$, we can take $a_{i j} \in A_{i} \cap A_{j}^{\prime}$.
It is easy to see that all these $a_{i j}$ are distinct (otherwise, if $a \in A_{i} \cap A_{j}^{\prime}$ and $a \in A_{s} \cap A_{t}^{\prime}$. Then $a \in A_{i} \cap A_{s}$. But when $i \neq s$, $A_{i} \cap A_{s}=\varnothing$, so only $i=s$. Similarly, $j=t$, a contradiction).
Therefore, $A_{1} \cup A_{2} \cup \cdots \cup A_{n}$ contains at least
$$
45(n-45) \geqslant 45^{2}>2014
$$

distinct elements, which is impossible.
Next, prove that when $n \leqslant 89$, for any $n$-vertex simple graph, there exist subsets $A_{1}, A_{2}, \cdots, A_{n}$ that satisfy the conditions.
Use mathematical induction to prove a more general conclusion:
For any $n$-vertex simple graph, there always exist $n$ subsets $A_{1}, A_{2}, \cdots, A_{n}$ of $\left\{1,2, \cdots, s_{n}\right\}$ that satisfy the conditions, where $s_{n} = \left[\frac{n^{2}}{4}\right]$ (when $n=1$, $A_{1}$ can only be the empty set).
When $n=1$, the condition is not contradictory, so the conclusion holds.
When $n=2$, let $A_{1}=\{1\}$, and determine $A_{2}$ to be $\{1\}$ or the empty set based on whether $V_{1}$ and $V_{2}$ are adjacent, so the conclusion still holds.

Assume the conclusion holds for $n=k$, and we need to prove it for $n=k+2$.

If no two vertices are adjacent, take all $A_{i}$ to be the empty set.

Next, assume there exist adjacent vertices, without loss of generality, let $V_{1}$ and $V_{2}$ be adjacent. By the induction hypothesis, for the induced subgraph formed by the other $k$ vertices $V_{i}$ $(3 \leqslant i \leqslant k+2)$, there exist $k$ subsets $A_{i}^{\prime}(3 \leqslant i \leqslant k+2)$ of $\left\{1,2, \cdots, s_{k}\right\}$ that satisfy the corresponding conditions. Let $A_{1}^{\prime}=A_{2}^{\prime}=\varnothing$.
Call positive integers greater than $s_{k}$ "new elements".
Since $V_{1}$ and $V_{2}$ are adjacent, take the new element $s_{k}+1$ and add it to $A_{1}^{\prime}$ and $A_{2}^{\prime}$.

For any $V_{i}(3 \leqslant i \leqslant k+2)$, if $V_{i}$ is not adjacent to both $V_{1}$ and $V_{2}$, no new elements are needed; if $V_{i}$ is adjacent to both $V_{1}$ and $V_{2}$, take the smallest unused new element and add it to $A_{1}^{\prime}$, $A_{2}^{\prime}$, and $A_{i}^{\prime}$; if $V_{i}$ is adjacent to one of $V_{1}$ and $V_{2}$, say $V_{1}$, take the smallest unused new element and add it to $A_{1}^{\prime}$ and $A_{i}^{\prime}$, but not to $A_{2}^{\prime}$. In any case, each $V_{i}(i \geqslant 3)$ uses at most one new element.
In total, at most $1+k$ different new elements are used.
After a series of operations of adding new elements, let $A_{1}^{\prime}, A_{2}^{\prime}, \cdots, A_{k+2}^{\prime}$ become $A_{1}, A_{2}, \cdots, A_{k+2}$. Then for any $i, j(1 \leqslant i<j \leqslant k+2)$, $A_{i} \cap A_{j} \neq \varnothing$ if and only if $V_{i}$ and $V_{j}$ are adjacent. Also, no more than $1+k$ new elements are used, so the largest element does not exceed $s_{k}+1+k=s_{k+2}$.
Thus, the conclusion holds for $n=k+2$.
Therefore, the conclusion holds for all positive integers $n$.
In particular, when $n \leqslant 89$, by
$$
s_{n} \leqslant 44 \times 45<2014 \text {, }
$$

there exist subsets $A_{1}, A_{2}, \cdots, A_{n}$ that satisfy the conditions.
In summary, the maximum value of $n$ is 89.