For all natural numbers $k$ the ratio $$ \frac{(2k)!}{(k!)^2}=\binom{2k}k $$ is an integer. From staring at the Pascal triangle long and hard, we know that these ratios grow rather quickly as $k$ increases. It is therefore natural to think that may be some factors from the numerator can be dropped in such a way that the ratio would still be an integer. More specifically, can we, for some carefully chosen $k$, leave out a chosen number of largest factors. In other words, given an integer $t>0$ does there exist a natural number $k$ such that $$\frac{(2k-t)!}{(k!)^2}\in\Bbb{Z}?$$

My curiosity about this comes from a question we had in May. The asker there had found the smallest $k$ that works for each of $t=1,2,\ldots,8$. In that question it was settled that with $t=9$ the smallest $k$ that works is $k=252970$.

It is natural to think about such divisibility questions one prime factor $p$ at a time. It is well known that if we write a natural number $m$ in base $p$, $$m=\sum_{i=0}^\ell m_ip^i$$ with the digits $m_i\in\{0,1,\ldots,p-1\}$, then the highest power of $p$ that divides $m!$ is equal to $$ \nu_p(m!)=\frac1{p-1}\left(m-\sigma_p(m)\right), $$ where $$\sigma_p(m)=\sum_{i=0}^\ell m_i$$ is the sum of "digits" of $m$ in base $p$. Written in this way, my question asks for a given $t$, whether there exists a $k$ such that the inequality $$ (2k-t)-\sigma_p(2k-t)\ge 2k-2\sigma_p(k) $$ holds for all primes $p\le k$.

As we have that slack one might expect this to be possible. But I'm not sure. One obstruction comes from primes just below $k$. If $k-(t/2)<p<k$, then $p^2$ is a factor in the denominator, but $2p$ is too large to appear as factor in the numerator, so $p^2\nmid (2k-t)!$. Occasionally a small prime is also problematic. It is not clear to me how to approach this. A construction may exist. The only thing this reminds me of is the elementary exercise $(k!)^{k+1}\mid (k^2)!$, but that doesn't seem to apply here.

In a comment under the answer to the linked question user metamorphy reports having confirmed this up to $t\le14$.

Edit/Note: The available evidence (see also Sil's comment under this question) suggests that, at least when looking for the smallest $k$ that works for a given $t$, whenever a chosen $k$ works for an odd number $t$, that same $k$ also works for $t+1$. If the main question proves to be too difficult to crack, steps towards explaining this phenomenon are also interesting.