Find all integers $k$ such that there exist 2017 integers $a_{1}, a_{2}, \cdots, a_{2017}$, satisfying that for any positive integer $n$ not divisible by the prime 2017, we have
$$
\frac{n+a_{1}^{n}+a_{2}^{n}+\cdots+a_{2017}^{n}}{n+k} \in \mathbf{Z} .
$$
First, we prove a lemma.
Lemma There exist infinitely many primes of the form $2017k + 1$.
Proof Assume there are only finitely many primes of the form $2017k + 1$, denoted as $p_{1}, p_{2}, \cdots, p_{n}$.
Let $m = 2017 p_{1} p_{2} \cdots p_{n}$,
$$
t = \frac{m^{2017} - 1}{m - 1} = m^{2016} + m^{2015} + \cdots + m + 1.
$$

Consider a prime factor $q$ of $t$. Then $q \nmid m$.
If $m \equiv 1 \pmod{q}$, then $q \mid 2017$.
Since $2017 \mid m$, it follows that $q \mid m$, which is a contradiction.
Thus, $m \not\equiv 0, 1 \pmod{q}$.
Therefore, $m^{2017} \equiv 1 \pmod{q}$.
By Fermat's Little Theorem, we know $m^{q-1} \equiv 1 \pmod{q}$.
Hence, $2017 \mid (q-1)$, which means $q$ is a prime of the form $2017k + 1$.
Since $q \nmid m$, it follows that $q \notin \{p_{1}, p_{2}, \cdots, p_{n}\}$, which is a contradiction.
The lemma is proved.
By the lemma, there exist infinitely many primes $p$ such that $p \equiv 1 \pmod{2017}$.
Take any such $p$.
By the Chinese Remainder Theorem, there exists a natural number $n$ such that
$$
\left\{\begin{array}{l}
n \equiv -k \pmod{p}, \\
n \equiv 1 \pmod{p-1}.
\end{array}\right.
$$

Then $n$ does not contain the prime factor 2017.
By Fermat's Little Theorem, we know $p \mid a(a^{p-1} - 1)$.
Since $(p-1) \mid (n-1)$, it follows that
$$
(a^{p-1} - 1) \mid (a^{n-1} - 1).
$$

Thus, $p \mid (a^n - a)$.
Notice that,
$$
\begin{array}{l}
p \mid (n + k), \\
(n + k) \mid (n + a_1^n + a_2^n + \cdots + a_{2017}^n).
\end{array}
$$
Hence,
$$
(n + k) \mid (a_1^n + a_2^n + \cdots + a_{2017}^n - k) \\
\Rightarrow p \mid (a_1^n + a_2^n + \cdots + a_{2017}^n - k) \\
\Rightarrow p \mid (a_1 + a_2 + \cdots + a_{2017} - k).
$$

By the arbitrariness of $p$, we know $p$ can be arbitrarily large.
Thus, $a_1 + a_2 + \cdots + a_{2017} - k = 0$.
By the Chinese Remainder Theorem, there exists a natural number $m$ such that
$$
\left\{\begin{array}{l}
m \equiv -k \pmod{p}, \\
m \equiv 2 \pmod{p-1}.
\end{array}\right.
$$

Then $m$ does not contain the prime factor 2017.
By Fermat's Little Theorem, we know $p \mid a(a^{p-1} - 1)$.
Since $(p-1) \mid (m-2)$, it follows that
$$
(a^{p-1} - 1) \mid (a^{m-2} - 1) \\
\Rightarrow p \mid a^2(a^{m-2} - 1) \\
\Rightarrow p \mid (a^m - a^2).
$$
By $p \mid (m + k)$,
$$
\begin{array}{l}
\Rightarrow p \mid (a_1^m + a_2^m + \cdots + a_{2017}^m - k) \\
\Rightarrow p \mid (a_1^2 + a_2^2 + \cdots + a_{2017}^2 - k).
\end{array}
$$

Since $p$ can be arbitrarily large, we have
$$
a_1^2 + a_2^2 + \cdots + a_{2017}^2 - k = 0.
$$

Thus, $a_1 + a_2 + \cdots + a_{2017}$
$$
= a_1^2 + a_2^2 + \cdots + a_{2017}^2.
$$

Since $a_i (i=1,2, \cdots, 2017)$ are integers, then $a_i^2 \geq a_i$. The equality holds if and only if $a_i = 0$ or 1.
Thus, $0 \leq k \leq 2017$.
When $k = m (0 \leq m \leq 2017)$, let $a_1, a_2, \cdots, a_{2017}$ have $m$ ones and $2017 - m$ zeros.
Hence, $k = 0, 1, \cdots, 2017$ satisfy the condition.