4. (Lokni Army) Let $A$ be a finite set, $N=\{1, 2, 3, \cdots\}$. If there exists a function $f: N \rightarrow A$ with the following property: if $|i-j|$ is a prime number, then $f(i) \neq f(j)$. Try to find the minimum number of elements in $A$.

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Consider in $N$ taking $i=1,3,6,8$. The absolute value of the difference between any two of these numbers is a prime number, so their images $f(1), f(3), f(6), f(8)$ are the same, meaning that the set $A$ must have at least 4 elements.

Below, we prove that the minimum number of elements in set $A$ is 4. For this, we construct a function $f$ that satisfies the given conditions for a four-element set.
$$
\begin{array}{c}
\text { Let } A=\left\{a_{0}, a_{1}, a_{2}, a_{3}\right\} \text { and } N_{k} \\
=\{x \mid x=k+4 n, n \in Z\}(k=0,1,2,3) .
\end{array}
$$

For any $x \in N$, define $f(x)=a_{k}$ if $x \in N_{1}$. In this case, for any $i, j \in N$, if $f(i) = f(j)$, by the definition of the function, $i$ and $j$ must be elements of the same set $N_{1}$, so $i-j$ is divisible by 4, and thus $|i-j|$ cannot be a prime number. This means that for a four-element set $A$, a function $f$ that satisfies the conditions exists. In summary, the minimum number of elements in set $A$ is 4.