Let $S:=\displaystyle\sum_{k=1}^{100}\,\dfrac1{k}$, and write $[n]:=\{1,2,\ldots,n\}$ for each positive integer $n$. Note that $B$ is not divisible by $5$ because
$$\begin{align}S&=\sum_{\substack{k\in[100]\\ 5\nmid k}}\,\dfrac1k +\frac15\,\sum_{\substack{k\in[20]\\5\nmid k}}\frac1k+\frac1{25}\,\sum_{k=1}^4\,\frac1k\tag{1}
\\&=\sum_{\substack{k\in[100]\\ 5\nmid k}}\,\dfrac1k +\frac15\,\sum_{t=0}^{3}\,\sum_{r=1}^4\frac1{5t+r}+\frac{1}{25}\cdot\frac{25}{12}\tag{2}
\\&=\sum_{\substack{k\in[100]\\ 5\nmid k}}\,\dfrac1k +\frac15\,\sum_{t=0}^{3}\,\left(\frac{10t+5}{(5t+1)(5t+4)}+\frac{10t+5}{(5t+2)(5t+3)}\right)+\frac{1}{12}\tag{3}
\\&=\sum_{t=0}^{19}\,\sum_{r=1}^4\,\frac{1}{5t+r} +\sum_{t=0}^{3}\,(2t+1)\left(\frac{1}{(5t+1)(5t+4)}+\frac{1}{(5t+2)(5t+3)}\right)+\frac{1}{12}\tag{4}\,.\end{align}$$
Now, $A$ is not divisible by $5$ because, from the equation above,
$$\begin{align}S&\equiv \sum_{t=0}^{19}\,\sum_{r=1}^4\,\frac{1}{5t+r}+\sum_{t=0}^3\,(2t+1)\left(\frac{1}{1\cdot 4}+\frac{1}{2\cdot 3}\right)+\frac{1}{12}\pmod{5}\tag{5}
\\&\equiv \sum_{t=0}^{19}\,\sum_{r=1}^4\,\frac{1}{r}+\left(\frac{1-5}{1\cdot 4}+\frac{1+5}{2\cdot 3}\right)\,\sum_{t=0}^3\,(2t+1)+\frac{1-25}{12}\pmod{5}\tag{6}
\\&\equiv \sum_{t=0}^{19}\,\frac{25}{12}+\big((-1)+1\big)(1+3+5+7)-2\pmod{5}\tag{7}
\\&\equiv 0+0\cdot 16-2 =-2\not\equiv 0\pmod{5}\tag{8}\,.\end{align}$$
Exercise for the Reader. Let $\dfrac{A}{B}=\sum\limits_{k=1}^{20}\,\dfrac{1}{k}$, where $A$ and $B$ are relatively prime positive integers. Prove that $5$ divides $A$ but $25$ does not divide $A$. Prove also that $5$ does not divide $B$.
