Inspired by Dev Sharma

First we note that for any even $k$ we have that $16|k^{5}.$

Next, we observe that $k^{5} - k = k(k^{4} - 1) = k(k^{2} - 1)(k^{2} + 1) = k(k - 1)(k + 1)(k^{2} + 1).$ For odd $k,$ each of $(k - 1), (k + 1)$ and $(k^{2} + 1)$ is divisible by $2,$ and precisely one of $(k - 1), (k + 1)$ is divisible by $4.$ We can thus conclude that for odd $k$

$k^{5} - k \equiv 0 \pmod{16} \Longrightarrow k^{5} \equiv k \pmod{16}.$

We can then say that the given sum $S$ is equivalent to the sum of all odd $k$ modulo $16,$ i.e.,

$S = \displaystyle\sum_{k=1}^{100} k^{5} \equiv \sum_{n=1}^{50} (2n - 1) \equiv 2*\dfrac{50*51}{2} - 50 \equiv 50^{2} \equiv (3*16 + 2)^{2} \equiv 2^{2} \pmod{16}.$

We thus can conclude that the remainder when $S$ is divided by $16$ is $\boxed{4}.$