To prove that there exist infinitely many natural numbers $n$ such that $n \mid 1^n + 2^n + \dots + k^n$, we will use a lemma and consider two cases based on the parity of $k$.

### Lemma:
If $p$ is an odd prime and $p \mid x + y$, then:
$$p^k \mid x^{p^k} + y^{p^k}$$
This lemma can be proven by induction on $k$.

### Proof of Lemma:
Base Case:
For $k = 1$, if $p \mid x + y$, then $x \equiv -y \pmod{p}$. Thus,
$$x^p + y^p \equiv (-y)^p + y^p \equiv -y^p + y^p \equiv 0 \pmod{p}$$
So, $p \mid x^p + y^p$.

Inductive Step:
Assume the lemma holds for $k = m$, i.e., $p^m \mid x^{p^m} + y^{p^m}$. We need to show it holds for $k = m+1$.

Consider $x^{p^{m+1}} + y^{p^{m+1}}$. We can write:
$$x^{p^{m+1}} + y^{p^{m+1}} = (x^{p^m})^p + (y^{p^m})^p$$
By the inductive hypothesis, $p^m \mid x^{p^m} + y^{p^m}$. Let $x^{p^m} + y^{p^m} = p^m z$ for some integer $z$. Then,
$$(x^{p^m})^p + (y^{p^m})^p = (p^m z)^p = p^{mp} z^p$$
Since $p$ is an odd prime, $mp \geq m+1$. Therefore,
$$p^{m+1} \mid p^{mp} z^p$$
Thus, $p^{m+1} \mid x^{p^{m+1}} + y^{p^{m+1}}$, completing the induction.

### Case 1: $k \equiv 0 \pmod{2}$
Let $p$ be an odd prime divisor of $k + 1$. Since $k$ is even, $p$ is an odd prime. By the lemma, for any $a > 0$,
$$p^a \mid x^{p^a} + (k - x)^{p^a} \quad \forall x = 1, 2, \ldots, \frac{k-1}{2}$$
Therefore,
$$p^a \mid \sum_{x=1}^k x^{p^a}$$
This means $n = p^a$ for any $a > 0$ satisfies the condition.

### Case 2: $k \equiv 1 \pmod{2}$
Similarly, let $p$ be an odd prime divisor of $k + 1$. By the lemma, for any $a > 0$,
$$p^a \mid x^{p^a} + (k - 1 - x)^{p^a}$$
and
$$p^a \mid k^{p^a}$$
Therefore,
$$p^a \mid \sum_{x=1}^k x^{p^a}$$
This means $n = p^a$ for any $a > 0$ satisfies the condition.

Thus, in both cases, we have shown that there exist infinitely many natural numbers $n$ such that $n \mid 1^n + 2^n + \dots + k^n$.

$\blacksquare$