1. **Define the Set $S$:**
$$S = \{n \in \mathbb{Z}_{>0} : \text{ for each prime } p \mid n \text{ and prime } q \mid (p-1) \text{ we have } q \mid n\}$$
We aim to show that $S = N$.

2. **Prove $S \subseteq N$:**
- Let $n \in S$ and $k$ be a positive integer such that $n \mid k^k - 1$. We need to show $n \mid k - 1$, i.e., $k \equiv 1 \pmod{n}$.
- Consider a prime $p$ such that $p^e \parallel n$ (i.e., $p^e \mid n$ and $p^{e+1} \nmid n$).
- Let $o$ be the order of $k$ modulo $p^e$. By Euler's theorem, $o \mid \varphi(p^e)$, and since $n \mid k^k - 1$, we have $o \mid k$, so $o \mid \gcd(k, \varphi(p^e))$.
- Since $\gcd(k, n) = 1$, it follows that $p \nmid k$. Therefore, $o \mid \gcd(k, p-1)$.
- If there is a prime $q \mid o$, then $q \mid \gcd(k, p-1)$, which implies $q \mid p-1$ and $q \mid k$. Since $n \in S$, we get $q \mid n$ and $q \mid k$, contradicting $\gcd(n, k) = 1$. Thus, $o = 1$ and $n \mid k - 1$.

3. **Prove $N \subseteq S$:**
- Assume for contradiction that there exists $n \in N$ such that $n \notin S$. Then there is a prime $p \mid n$ and a prime $q \mid p-1$ such that $q \nmid n$.
- Let $v > 0$ such that $p^v \parallel n$. Since $p$ and $q$ are primes and $q \nmid n$, $q$, $n/p^v$, and $p^v$ are pairwise coprime.
- By the Chinese Remainder Theorem, we can choose a positive integer $k$ such that:
$$\begin{cases} k \equiv 0 \pmod{q} \\ k \equiv 1 \pmod{n/p^v} \\ k \equiv g^{\varphi(p^v)/q} \pmod{p^v} \end{cases}$$
where $g$ is a primitive root modulo $p^v$.
- Since $g$ is a primitive root and $\frac{\varphi(p^v)}{q} < \varphi(p^v)$, it follows that $k \equiv g^{\varphi(p^v)/q} \not\equiv 1 \pmod{p^v}$, so $p^v \nmid k - 1$, implying:
$$n \nmid k - 1 \tag{1}$$
- Since $q \mid k$, we can write $k = \ell q$ for some positive integer $\ell$. By Euler's theorem:
$$k^k = \left( g^{\varphi(p^v)/q} \right)^{\ell q} = (g^q)^{\varphi(p^v)} \equiv 1 \pmod{p^v}$$
and since $k^k \equiv 1^k \equiv 1 \pmod{n/p^v}$, we get $k^k \equiv 1 \pmod{n}$, implying $n \mid k^k - 1$. From this and (1), it follows that $n \notin N$, a contradiction. Therefore, $S = N$.

4. **Show $\gcd(n_1, n_2) \in S$ for $n_1, n_2 \in S$:**
- Let $p \mid \gcd(n_1, n_2)$ and $q \mid p-1$. Since $p \mid n_i$ and $n_i \in S$, we get $q \mid n_i$. Thus, $q \mid \gcd(n_1, n_2)$. Therefore, $\gcd(n_1, n_2) \in S$.

$\blacksquare$