Integer Values of $\sum_{k=1}^n k^r . \sum_{q=1}^n \frac{1}{q^r}$ For harmonic numbers $H_n = \sum_{k=1}^n \frac{1}{k}$ we know that this sum is never integer for any $n$. The same is true for generalized Harmonic numbers: the sum $\sum_{k=1}^n\frac{1}{k^r}$ is never integer for any $n$ obviously as it is alwas smaller than $2$ for any $r$.
Now looking for this sum product integer values:
$$\sum_{k=1}^n k^r \cdot\sum_{q=1}^n \frac{1}{q^r}$$
The only non-trivial ($n\ne1$) integer value I have found is $11$ for $r=1$ and $n=3$.  So  $(1+2+3)(\frac{1}{1}+\frac{1}{2}+\frac{1}{3})=11$
Or we can use other notation instead of sums:
$$H_{n,r}\cdot H_{n,-r}$$
$H_{n,r}$ is a generalized harmonic numbers, see here.
Just examined the other values $n<1000$ and $r\le4$, but never get any integer value.
So the question is "Is the above mention sum product has integer values except the trivial ones and above mentioned 11?" I can not find a clue to solve this task.

Have checked for $r<10$ and $n<1000$. No other integer except this 11 wsa found.

After getting all integer values for $r<10$, $n<1000$, $m<1000$ of $H_{n,r} \cdot H_{m,-r}$ (excel file is here) found that usually $n$ is strongly greater than $m$. The only exception is case with $n=m$ for above mentioned 11 case. So seems there would be a formula about the limit of $n$ depending on $r$ and $m$ for the cases when this product is integer.

Edit: please refer to the proof of the conjecture in the OP in the end of this post.
Original post:
This is by no means a complete answer to the question, it is just an extension of the solutions provided in the OP.
Due to some confusion in the preliminary statement of the OP I have incorrectly understood that we are looking for integer values of the quantity (thanks to @Gevorg Hmayakyan for pointing this out)
$$p(r,n) = H_{n} H_{n,-r} \tag{1}$$
while it was in fact the more sophisticated quantity
$$q(r,n) = H_{n,r} H_{n,-r} \tag{2}$$
Here $H_n = \sum_{k=1}^{n}\frac{1}{k}$, $H_{n,-r} = \sum_{k=1}^{n} k^r$.
Results for $(1)$
It might be neverthess interesting to look at the integer solutions for $(1)$.
I have found so far for the tripel $\{r,\{n,p\}\}$ with $n\le 10^4$ the following solutions (beside the trivial solutions $\{r, \{1, 1\}\}$
$$\begin{array}{l}
 \{1,\{\{3,11\}\}\} \\
 \{2,\{\{7,363\}\}\} \\
 \{3,\{\{3,66\}\}\} \\
 \{4,\{\}\} \\
 \{5,\{\{3,506\}\}\} \\
 \{6,\{\}\} \\
 \{7,\{\{3,4246\}\}\} \\
 \{8,\{\}\} \\
 \{9,\{\{3,37026\}\}\} \\
 \{10,\{\{7,917393565\}\}\} \\
 \{11,\{\{3,328526\}\}\} \\
 \{12,\{\}\} \\
 \{13,\{\{3,2937946\}\}\} \\
 \{14,\{\}\} \\
 \{15,\{\{3,26366406\}\}\} \\
 \{16,\{\}\} \\
 \{17,\{\{3,236997266\}\}\} \\
 \{18,\{\}\} \\
 \{19,\{\{3,2131773886\}\}\} \\
 \{20,\{\}\} \\
\end{array}$$
We observe a strong preference for $n=3$ for serveral $r$ where we have $H_3=\frac{11}{6}$, and a surprising exception for $r=10$ with $n=7$..
Results for $(2)$
For $r=2$ and $n\le 10^4$ the smallest difference of $p$ to an integer is $d=0.000118...$ which is reached already at $n=1891$
Sketch of a proof of impossibility
It seems that for $r\ge 2$ (which we assume henceforth) the quantity $q(r,n)$ is never an integer for any $n\ge 2$.
The idea is to count powers of 2 in numerator and denominator and show that they don't cancel.
For $n\ge 2$ we have
$$H_{n,r}=\frac{N}{D}=\frac{o}{2^r i}\tag{3}$$
where $o$ is odd and $i$ is integer.
Hence for odd $H_{n,-r}$ the product $q=H_{n,r} H_{n,-r}=\frac{o}{e} \times o = \frac{o}{e} $ can't be integer because the factor $2$ in the denominator remains uncancelled.
Now what can be said about the parity of $H_{n,-r}$? There is a periodic pattern: for $n=1,2,3,4$ etc. is $ooee$ etc.
Hence writing
$$n=\frac{1}{2}(4 k-1 \mp 1)\tag{4}$$
which for $k=1,2,3,...$ covers all values of $n$, the parity of $H_{n,-r}$ is equal to the parity of $k$.
We conclude that for odd $k=2m-1$ i.e. $n=\frac{1}{2}(8m -5 \mp 1)$ the sum $H_{n,-r}$ is odd and hence $q$ is not an integer.
On the other hand, if $n=\frac{1}{2}(8m -1 \mp 1)$ the sum $H_{n,-r}$ is even and we can't be sure that the 2's don't cancel to lead to an integer product $q$.
But we still have not yet considered higher powers of two.
I did that now with the result that the power $s$ of $2$ in the denominator of $H_{n,+r}$ is always greater than the power $t$ of $2$ in $H_{n,-r}$. Hence their product can't be an integer.
This completes the proof for $r\ge 2$.
Q.E.D.
Remark: this proof needs to be presented more strictly, of course. But I am sure it is valid.
It can be shown that the power of 2 in the denominator of $H_{n,r}$ is given by
$$f_{H+}(n,r) = r (\left \lceil {\frac{\log(n+\epsilon)}{\log(2) }} \right \rceil-1)$$
where $\epsilon$ is an arbitrary small positive number.
