For $n \equiv 0, 1, 2 \pmod 9$, write $\{\,1,\dots,3n\,\}$ as the disjoint union of arithmetic progressions $A_1, A_2,\dots,A_n$ of length 3, where $A_i$ has step $i$.
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$\begingroup$ See mathoverflow.net/questions/61744 to get an idea of a solution. This is the wrong forum for your question. Gerhard "Trying Small Examples Helps Much" Paseman, 2017.04.26. $\endgroup$ – Gerhard Paseman Apr 26 '17 at 22:42

4$\begingroup$ I'm not sure whether this is the right forum or not, but I would like to know the source of this question, and what you have found in your own efforts to solve it, William. $\endgroup$ – Gerry Myerson Apr 26 '17 at 23:29

1$\begingroup$ Wait, so how does this work for $n=2$? $\endgroup$ – RP_ Apr 27 '17 at 0:13

1$\begingroup$ I tried it for some small values (small but bigger than 2) and it is possible to do this for n = 9, 10, 11, 18, 19. 20. For 10, for example, the following works: (1, 2, 3) (4, 6, 8) (7, 10, 13) (20, 24, 28) (11, 16, 21) (17, 23, 29) (12, 19, 26) (14, 22, 30) (9, 18, 27) (5, 15, 25) $\endgroup$ – Wade HannCaruthers Apr 28 '17 at 19:33

1$\begingroup$ Curiously, it appears not to work for numbers that do not satisfy your assumption, suggesting you may have some obstruction for those...care to share? :) $\endgroup$ – Wade HannCaruthers Apr 28 '17 at 19:48