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Proving the integrality of an fractions of factorials can be done through De Polignac formula for the exponent of factorials, reducing the question to an floored inequality.
Some of those inequalities turn out to be very hard to proof if true at all.
The first is, given $x_i in mathbb{R}$ and ${x_i} = x_i - lfloor x_i rfloor$:

$$sum_{i=1}^{n}left lfloor n {x_i} right rfloor geq left lfloor sum_{i=1}^{n}{x_i} right rfloor$$

I was able to prove this one by arguing that if $left lfloor sum_{i=1}^{n}{x_i} right rfloor = L$ than there is some $x_k geq frac{L}{n}$, so the left side is at least $L$. But I was unable to apply the same idea to the following inequality:

$$sum_{i=1}^{n}left lfloor q_i {x_i} right rfloor geq left lfloor sum_{i=1}^{n}{x_i} right rfloor$$
Where $q_i in mathbb{N}$ and $frac{1}{q_1} + dotsm + frac{1}{q_n} leq 1$.
Also, this generalization was proposed:
$$sum_{i=1}^{n}left lfloor q_i {x_i} right rfloor geq left lfloor sum_{i=1}^{n}k_i{x_i} right rfloor$$
Where $q_i, k_i in mathbb{N}$ and $frac{k_1}{q_1} + dotsm + frac{k_n}{q_n} leq 1$.
I don't know if the last two inequalities are correct neither know how to proof if wrong or any counter-example if not. Could someone help?

Let $theta_i={x_i}$, so that the second inequality reads
$$sum_{i=1}^{n}left lfloor q_i theta_i right rfloor geq left lfloor sum_{i=1}^{n}theta_i right rfloor. tag{1}$$
Also let $L$ denote the right side, as in the proof of the OP.

Now if for each $i$ we had $theta_i<L/q_i$ then we would have
$$sum_{i=1}^n theta_i<L sum_{i=1}^n frac{1}{q_i} le L,$$

the last inequality from the assumption that the reciprocals of the $q_i$ sum to at most $1$. From this, similar in spirit to the OP's proof, we get that for at least one index $i$ we have $theta_i ge L/q_i$, in other words $q_i theta_i ge L$, implying that the term $left lfloor q_i theta_i right rfloor ge L$ and establishing (1).

Perhaps a similar idea would work for the final inequality.

ADDED: Yes, the third inequality has a similar proof. With the notation above it reads

$$sum_{i=1}^{n}left lfloor q_i theta_i right rfloor geq left lfloor sum_{i=1}^{n}k_itheta_i right rfloor. tag{2}$$
Again let $L$ denote the right side, and assume for each $i$ we had $theta_i<L/q_i$. then we would have
$$sum_{i=1}^n k_itheta_i<L sum_{i=1}^n frac{k_i}{q_i}le L,$$
using $sum (k_i/q_i) le 1.$
This as before implies there is an index $i$ for which $q_i theta_i ge L$ to finish.