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.
 A: 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.
A: For the case $r=1$, $n=3$ is (I believe*) provably the only solution for $H_{n,1}H_{n,-1} = A_n \in \mathbb{Z}$.
Things we can note:

*

*The sequence $H_{n,1}$ is the same as the triangular numbers $T_n=\frac12(n^2-n)$, so $T_n \in O(n^2)$

*The sequence $H_{n,-1} (=H_n)$ are elements of $\mathbb{Q}$, so we can write each as $\frac{c_n}{d_n}$, in lowest terms of course. For instance, for $H_5$, $c_n = 137$ and $d_n = 60$.

*Both $c_n$ and $d_n$ are mostly increasing sequences, though they have ups and downs.

*The value of $d_n$ has, as a maximum, $L_n = \text{lcm} (1,2, \cdots n-1, n)$. This is an increasing function, but $d_n < L_n$ most of the time.

We can safely say:
$$A_n \in \mathbb{Z} \implies d_n \mid T_n$$
If $T_n$ doesn't cancel the denominator completely, our product cannot be an integer. This further implies $T_n \geq d_n$, as otherwise $d_n$ cannot be a divisor of $T_n$. To prove that $3$ is the only solution, it suffices to show that $T_n < d_n$ for all $n \geq 6$. By inspection, $1,2,4$ and $5$ are not solutions, and of course $3$ is. We can mostly worry about $d_n$ from here on.
For $n=6$, $d_n = 20$, and $T_n=21$. While $T_n > d_n$, obviously they don't cancel one another, so $6$ is not a solution. Now let's consider what the next fraction in the sequence is, for a given $n$:
$$H_{n+1} = H_n + \frac{1}{n+1} = \frac{c_n}{d_n} + \frac{1}{n+1} = \frac{(n+1)c_n + d_n}{(n+1)d_n} = \frac{c_{n+1}}{d_{n+1}}$$
If $(c_{n+1}, d_{n+1}) = 1$, then this fraction is in lowest terms, and the $d_n$ has been multiplied by $n+1$. This happens at every prime number:
$$H_{p} = \frac{c_{p-1}}{d_{p-1}} + \frac{1}{p} = \frac{pc_{p-1} + d_{p-1}}{pd_{p-1}} = \frac{c_p}{d_p}$$
Because the original fraction was in lowest terms, $c_{p-1}, d_{p-1},$ and $p$ share no common factors. The sum of two coprime integers, if composite, cannot share any factors with the original integers. But $pc_{p-1}$ and $d_{p-1}$ both share a common factor with $pd_{p-1}$, meaning the sum $c_p = pc_{p-1} + d_{p-1}$ has no common factors with $d_p$. Hence $(c_p, d_p)=1$ and this fraction cannot be reduced further.**
At multiples of primes, we have a more interesting result. Setting $m = kp-1$:
$$H_{kp} = \frac{c_m}{d_m} + \frac{1}{kp} = \frac{kpc_m +d_m}{kpd_m}$$
Now, either $d_m$ has a factor in common with $p$, or it doesn't. If it doesn't, then we have no further to discuss. If it does, then $d_m = pq$, and we can cancel $p$ from the fraction. We may even have $d_m = kpq$, in which case the denominator remains the same. For example, $d_{10} = d_9$, and in fact this will be the case for every $2p$, as $d_n$ is always even.
We haven't yet examined the addition in the numerator, however. After cancelling $p$ or $kp$, the two addends share no factors with each other, but also not with the denominator. Their sum, however, might have other factors in common with the denominator. When this happens, the denominator decreases--this happens, for instance, at $n= 6,18,20,21,33$ for $p = 3,3,5,3,11$. In this way, we can have $d_5 = 60$ and $d_6 = 20$.
But... in the same situation, where we are adding $\frac{1}{kp}$, and $d_{n-1}$ does not have a factor of $p$, we find that $d_n = pd_{n-1}$. These increases and decreases happen in patterns that are tough to describe, but it is obvious that:

*

*Each time $n$ is a prime, $d_n = pd_{n-1}$

*When a prime factor has been fully removed from the denominator, it will eventually be replaced, reversing the removal.

*This process means that usually, possibly always, $d_n \geq \frac{L_n}{n^2}$. (It's conjectured that $L_n = d_n$ infinitely many times.)

Hence, $d_n$ is increasing, and $d_n \sim O(L_n)$. This growth is much, much higher than $O(n^2)$. Therefore there can be no solutions $n: H_{n,1}H_{n,-1} \in \mathbb{Z}$ above $n=6$.

Interestingly (by inspection), it appears that
$$L_n = n\# \prod_{p \ \text{prime}}^{p \leq \sqrt{n}} p( \lfloor \log_p n \rfloor -1)$$
That is, $\text{lcm}(1,2,3,\cdots n)$ gains a factor of $n$ anytime $n = p^k, k \geq 2$.

*(A review of the proof will likely find holes.)
**(Note: By inspection, this also happens at prime powers, but the multiplier is always $2$ for $2^k$, is $3$ itself and $9$ for all other $3^k$, and for all other primes appears that $p$ and $p^2$ are the multipliers for themselves, and all other $p^k$ causes multiplication by $p^3$. I haven't attempted to prove this yet.)
