$\sum n^k$

We will show that all odd integers between 1 and 100 inclusive work, and only these values work. Indeed, if $k$ were even, then we can take $n = 2$ as a counterexample.

Suppose $k$ were an odd integer. Let $N = \dfrac{1^k + 2^k + 3^k + \cdots + n^k}{1 + 2 + 3 + \cdots + n}.$ We want to show that $N$ is an integer. Using the formula for the sum of the first $n$ positive integers, we get

$N = \dfrac{1^k + 2^k + 3^k + \cdots + n^k}{\frac{n(n + 1)}{2}} = \dfrac{2(1^k + 2^k + 3^k + \cdots + n^k)}{n(n + 1)}.$

Now, we split into cases based on the parity of $n$ :

Case 1: $n = 2a,$ for some positive integer $a$

We have

$N = \dfrac{2(1^k + 2^k + 3^k + \cdots + (2a)^k)}{2a(2a + 1)} = \dfrac{1^k + 2^k + 3^k + \cdots + (2a)^k}{a(2a + 1)}.$

Since $k$ is odd, we have $p^k + ((2a + 1) - p)^k \equiv 0 \! \pmod{2a + 1}$ and $p^k + (2a - p)^k \equiv 0 \! \pmod{a},$ for all positive integers $p$ between 1 and $2a$ inclusive. Thus,

$1^k + 2^k + 3^k + \cdots + (2a)^k \equiv (1^k + (2a)^k) + (2^k + (2a - 1)^k) + \cdots + (a^k + (a + 1)^k) \equiv 0 \! \! \! \! \pmod{2a + 1},$

and

$1^k + 2^k + 3^k + \cdots + (2a)^k \equiv a^k + (2a)^k + (1^k + (2a - 1)^k) + (2^k + (2a - 2)^k) + \cdots + ((a - 1)^k + (a + 1)^k) \equiv 0 \! \! \! \! \pmod{a},$

so we conclude that $N$ is an integer.

Case 2: $n = 2a + 1$ for some nonnegative integer $a$

We have

$N = \dfrac{2(1^k + 2^k + 3^k + \cdots + (2a + 1)^k)}{(2a + 1)(2a + 2)} = \dfrac{1^k + 2^k + 3^k + \cdots + (2a + 1)^k}{(a + 1)(2a + 1)}.$

Similar to the $n$ even case, we have $p^k + ((2a + 2) - p)^k \equiv 0 \! \pmod{a + 1}$ and $p^k + ((2a + 1) - p)^k \equiv 0 \! \pmod{2a + 1},$ for all positive integers $p$ between 1 and $2a + 1$ inclusive. Thus,

$1^k + 2^k + 3^k + \cdots + (2a + 1)^k \equiv (a + 1)^k + (1^k + (2a + 1)^k) + (2^k + (2a)^k) + \cdots + (a^k + (a + 2)^k) \equiv 0 \! \! \! \! \pmod{a + 1},$

and

$1^k + 2^k + 3^k + \cdots + (2a + 1)^k \equiv (2a + 1)^k + (1^k + (2a)^k) + (2^k + (2a - 1)^k) + \cdots + (a^k + (a + 1)^k) \equiv 0 \! \! \! \! \pmod{2a + 1},$

so $N$ is an integer.

Therefore, $1 + 2 + 3 + \cdots + n|1^k + 2^k + 3^k + \cdots + n^k$ when $k$ is odd. This gives us a final answer of $1 + 3 + 5 + \cdots + 99 = \boxed{2500}.$