Finding a Sum in 2019.

The number $\frac{2019!}{i!j!k!l!}$ represents the number of ways to separate a set $X$ with 2019 elements into a partition of four distinct subsets, $X_1$ , $X_2$ , $X_3$ , and $X_4$ , such that $|X_1| = i, |X_2| = j, |X_3| = k$ , and $|X_4| = l$ .

As we sum over possible values $i,j,k$ , and $l$ , we count the number of functions from the set $X$ to the set $\{1,2,3,4\}$ .

The number of such functions is $A=4^{2019}$ .

Then $\ln A = 2019 \ln 4 = 2798.928$ , and $\lfloor{\ln A}\rfloor = 2798$

It can be proved that $(x_1+x_2+x_3+x_4)^n$ is equal to the sum of all numbers of the form $\frac{n!}{i!j!k!l!} x_1^ix_2^jx_3^kx_4^l$ for all non-negative integers satisfying that $i+j+k+l=n.$ Then making $x_1=x_2=x_3=x_4=1,$ we obtain that the sum of all numbers of the form $\frac{n!}{i!j!k!l!}$ for all non-negative integers satisfying that $i+j+k+l=n,$ is equal to $4^n.$ Making $n=2019,$ we get that the number $A=4^{2019}.$ Then $\ln A = 2019 \ln 4 =2798. 92... .$ Therefore, the answer to this problem is $\boxed{2798}.$