Why does there not exist a particular finite set of congruences that forms a partition of the set of integers There is a paragragh that I saw in an article and could not really understand. The article is Covering Systems of Congruences, J. Fabrykowski and T. Smotzer, Mathematics Magazine Vol. 78, No. 3 (Jun., 2005), pp. 228-231, DOI: 10.2307/30044163. It starts by discussing problem 9 from the 2002 American Invitational Mathematics Examination (AIME):

PROBLEM. Harold, Tanya, and Ulysses paint a very long fence. Harold starts with the first picket and paints every $h$th picket; Tanya starts with the second picket and paints every $t$th picket; and Ulysses starts with the third picket and paints every $u$th picket. If every picket gets painted exactly once, find all possible triples $(h,t,u)$.

After presenting the solution, the article goes on to say:

One can generalize the AIME problem and ask whether there exists a finite set of congruences, with all moduli distinct and greater than or equal to $2$, that forms a partition of the set of integers. This turns out to be impossible. Relaxing the assumption about partitioning the integers, one can look for finite sets of congruences such that every integer belongs to at least one of them."

Can anyone help explain why the first part is impossible?

The answer is yes.  There are several.
This question gives one example: Prove {0 mod 2, 0 mod 3, 1 mod 4, 1 mod 6, 11 mod 12} is a covering system
$0 \mod 2, 0 \mod 3, 1\mod 4, 1\mod 6, 11 \mod 12$.
This paper is all about them https://arxiv.org/pdf/1705.04372.pdf
