Four, (50 points) Let $k$ be a given positive integer. Find all positive integers $a$ such that there exists $n \in \mathbf{N}_{+}, n^{2} \mid \left(a^{n}-1\right)$, and $n$ has exactly $k$ distinct prime factors.
First, we present a lemma.

Lemma: If $p \nmid (a-1)$, then $\left(a-1, \frac{a^{p}-1}{a-1}\right)=p$.
Proof is omitted.
Returning to the original problem.
When $a=1$, it clearly satisfies the condition.
First, we prove: $n \times\left(2^{n}-1\right) (n \geqslant 2)$.
Let $p$ be the smallest prime factor of $n$.
If $n \mid\left(2^{n}-1\right)$, then $p \mid\left(2^{n}-1\right)$.
By Fermat's Little Theorem, $p \mid\left(2^{p-1}-1\right)$.
Thus, $p \mid\left(2^{(n, p-1)}-1\right)$.
Since $(n, p-1)=1$, then $p \mid 1$, which is a contradiction.
Therefore, $a \neq 2$.
Next, we use mathematical induction to prove: When $a \geqslant 3$, the condition is satisfied.
When $k=1$, take $p \mid (a-1)$, by the lemma we know
$$
p^{2} \mid\left(a^{p}-1\right),
$$

Taking $n=p$ satisfies the condition.
Assume that for $k=s (s \geqslant 1)$, there exists $n^{2} \mid\left(a^{n}-1\right)$, where $n$ has exactly $s$ distinct prime factors, and $a^{n}-1$ has a prime factor $q$ that is coprime with $n$.
Then, by $\left(\frac{a^{n q}-1}{a^{n}-1}, a^{n}-1\right)=q$, we get $q \mid\left(a^{n q}-1\right)$.
Thus, $(n q)^{2} \mid\left(a^{n q}-1\right)$.
At this point, since $\frac{a^{n q}-1}{a^{n}-1}>q$, $a^{n q}-1$ must have a prime factor $p$ that is coprime with $n q$.

Since $n q$ has exactly $s+1$ distinct prime factors, for all $k \in \mathbf{N}_{+}$, there exists $n \in \mathbf{N}_{+}$ such that $n^{2} \mid\left(a^{n}-1\right)$, $n$ has exactly $k$ distinct prime factors, and $a^{n}-1$ has a prime factor $q$ that is coprime with $n$.

Therefore, the values of $a$ that satisfy the condition are all positive integers except 2.
(Li Jianguo, First Affiliated High School of Central China Normal University, 430000)