$\underline{\text { Folklore }}$

Does there exist such an $N$ and $N-1$ infinite arithmetic progressions with differences $2,3,4, \ldots, N$, such that every natural number belongs to at least one of these progressions?
Let $N=12$. Each natural number can be written in the form $12 k+r$, where $r$ is one of the numbers $0,1, \ldots, 11$. All numbers for which $r$ is even belong to the progression $2,4,6, \ldots$; all numbers for which $r$ is a multiple of $3$ belong to the progression $3,6,9, \ldots$ The remaining numbers have $r=1,5,7$ and 11. They are covered by the progressions $4 k+1,6 k+7$ and $12 k+$ $11(k=0,1, \ldots)$ with differences 4,6 and 12.