Divisibility of subsets of the set $1, 2, 3, ..., n$

Question

Let $n$ be an even positive integer. Can one divide the numbers $1, ..., n$ into three nonempty groups, so that the sum of numbers in the first group is divisible by $n + 1$, in the second one by $n + 2$, and in the third one by $n + 3?$ For which odd integers would this be true?

Here is what I have so far:

1, 2, 3, 4, 5, 6, ... , n are the congruence classes for (n+1). So as n is even, we could re-write the classes as:

$1, 2, 3, 4,..., n/2, -n/2, -(n/2)-1, ..., -1 \equiv 0 \mod (n+1)$

$1, 2, 3, 4,..., (n+2)/2, -(n+2/2)-1, ..., -2 \equiv 2+(n/2) \mod (n+2)$

$1, 2, 3, 4,..., (n+2)/2, -(n+2/2), -(n+2/2)-1,..., -3 \equiv 2 \mod (n+3)$

Would this always be true? If not, how else would I approach the problem?

Answer 1

$$1+2+\cdots+(2n+1)=\frac{(2n+1)(2n+2)}2\\=(2n+2)+(n-3)(2n+3)+2(2n+4)$$ Set aside $1+(2n+1)$ for the first, and $4+5+(2n-1)+2n$ for the last.

Answer 2

This is a complete answer for even $n$ and a partial answer for odd $n$. (Added later: See Michael's answer for a complete solution of the odd $n$ case.)

Suppose we've partitioned the numbers $\{1,2,\ldots,n\}$ into three nonempty groups so that the sum of the numbers in the groups are $(a+1)(n+1)$, $(b+1)(n+2)$, and $(c+1)(n+3)$, with $a,b,c\ge0$. (Thus use of $(a+1)$, $(b+1)$ and $(c+1)$ instead of just $a$, $b$ and $c$ ensures nonemptiness.) Then

$$1+2+\cdots+n={n(n+1)\over2}=(3n+6)+a(n+1)+b(n+2)+c(n+3)$$

We can rewrite this as

$${n^2-5n-12\over2}=a(n+1)+b(n+2)+c(n+3)$$

Since $a,b,c\ge0$, we have

$$(a+b+c)(n+1)\le a(n+1)+b(n+2)+c(n+3)\le (a+b+c)(n+3)$$

This translates into

$${n^2-5n-12\over n+3}\le2(a+b+c)\le{n^2-5n-12\over n+1}$$

which can be rewritten as

$$n-8+{12\over n+3}\le2(a+b+c)\le n-6-{6\over n+1}$$

Setting aside the cases $n\le9$ for later, we see that these inequalities are satisfied if and only if

$$2(a+b+c)=n-7$$

and thus no even $n$ greater than $8$ can possibly satisfy the desired conditions.

The cases $n\le8$ are easily dispensed with as well. For example, for $n=8$, there is no even number between $12/11$ and $4/3$. To summarize:

There is no even positive integer that satisfies the OP's condition.

This leaves the OP's question as to what happens for odd $n$ (greater than $7$). My only observation here is to note that for $n=9$, we have

$$\begin{align} 1+9&=1\cdot10\\ 3+8&=1\cdot11\\ 2+4+5+6+7&=2\cdot12 \end{align}$$

(but again, see Michael's answer for general odd $n\ge9$).