Counting tuples of highly composite numbers whose product is also highly composite I have little experience in number theory. While messing around with highly composite numbers, I came up with the following problem, which I've been unable to solve:

Let $a, b, c, d$ be a quadruplet of highly composite numbers greater-than 1. Do there exist infinitely many of these quadruplets such that the product $abcd$ is also highly composite?

It seems natural to ask about general $k$-tuples with $k>1$. I strongly suspect that it holds for pairs, because I think there are infinitely many highly composite numbers $N$ such that $2N$ is also highly composite, but I haven't been able to prove or find existing results for even this much.
Is this a well-known or relatively easy problem? Do we know for what $k$ this holds? If it's known to be finite, is there a known expression for how many such tuples there are?
 A: Just the case of multiplying by 2. For any integer $k \geq 3,$  construct positive real number
$$ \delta = \frac{\log \left(1 + \frac{1}{k} \right)}{\log 2}.  $$
Then construct the superior highly composite number
$$ W_\delta = \prod_p \; p^{\left\lfloor \frac{1}{p^\delta - 1} \right\rfloor  } $$
Then $W_\delta $ is an even integer and half of it is also an SHC  number.
Therefore, take your $$ N = \frac{1}{2}  W_\delta \; \; , $$
both $N$ and $2N$ are highly composite.
A: Here's my attempt to fill in a justification for Will's claim that "half of it is also an SHC number":
Note that $\left\lfloor\frac{1}{p^{\delta}-1}\right\rfloor = 0$ for  $p > 2^{\frac{1}{\delta}}$, and so we only consider primes $2 < p \le 2^{\frac{1}{\delta}}$. Since $\frac{1}{p^{\delta}-1}$ is an integer iff $p=2$, there must exist some $\epsilon\in(0,1)$ such that for $2 < p \le 2^{\frac{1}{\delta}}$ we have $\left\{\frac{1}{p^{\delta}-1}\right\} > \epsilon$, where $\{x\}$ denotes the fractional part of $x$. To construct $N$ such that $N=\frac{1}{2}W_{\delta}$, we choose some ​$c>0$ such that for all $p \le 2^{\frac{1}{\delta}}$ we have
$$\frac{1}{p^{\delta}-1} - \frac{1}{p^{\delta+c}-1}<\epsilon$$
Now, note that $\left\lfloor\frac{1}{2^{\delta+c}-1} \right\rfloor = \left\lfloor\frac{1}{2^{\delta}-1}\right\rfloor - 1$, and for $p>2$ we have $\left\lfloor\frac{1}{p^{\delta+c}-1} \right\rfloor = \left\lfloor\frac{1}{p^{\delta}-1} \right\rfloor$. Hence
$$N=\prod_p p^{\left\lfloor\frac{1}{p^{\delta+c}-1} \right\rfloor} = \frac{1}{2}W_{\delta}$$
