Divide the first 5n positive integers into $5$ groups such that the sum of the numbers in each group is equal Prove that for every integer $ n> 1$, it is possible to divide the first 5n positive integers into $5$ groups such that each group has exactly $n$ numbers and the sum of the numbers in each group is equal.
If n is even, I divide into 5 groups as follows:
$A_1: 1,3,5,...,n-1,4n+2,4n+4,…,5n$
$A_2: 2,4,6,...,n,4n+1,4n+3,…,5n-1$
$A_3: n+1,n+3,n+5,...,2n-1,3n+2,3n+4,…,4n$
$A_4:  n+2,n+4,...,2n,3n+1,3n+3,…,4n-1$
$A_5:  2n+1,2n+2,…,3n$
$\Rightarrow A_1=A_2=A_3=A_4=A_5=\frac{5n^2+n}{2}$
If n is odd, I don't know how to prove. Please help me.
 A: Whenever $p$ is some prime, every group of $pn$ ($n > 1$) integers can be split into $p$ groups such that the sum of numbers in each group is equal. Unfortunately, this proof is only existential.
$\textbf{Proof;}$
Set $S_{p,n} = \{1,...,pn\}$ and set $F_{n,p} = (2^{S_{p,n}})^{p}$ so that $F_{n,p}$ is the set of $p$-tuples where each tuple entry is a subset of $S_{p,n}$. We will say that an element $x$ of $F_{n,p}$ is valid whenever we have
$$\bigcup_{j=1}^{N}(x)_{j} = S_{p,n}$$
$$(x)_{i} \cap (x)_{j} = \emptyset \text{ whenever }i \neq j$$
here $(x)_{j}$ is the projection of $x$ onto its $j-$th tuple component. We will denote the set of valid elements of $F_{n,p}$ as $F_{n,p}^{*}$. Given a finite set $A$ such that $A \subset \mathbb{N}$ we will denote
$$\text{Sum}(A) = \sum_{y \in A}y.$$
We define $\phi_{n,p}: F_{n,p}^{*} \rightarrow \mathbb{C}$ via
$$\phi_{n,p}(x) = \sum_{j=0}^{p-1}\text{Sum}((x)_{j+1})e^{\frac{2\pi i j}{p}}$$
Note that
$$\phi_{n,p}(x) = 0 \text{ iff } \text{Sum}((x)_{1}) = \text{Sum}((x)_{2})...= \text{Sum}((x)_{p})$$ due to Eisenstein's criterion applied to cyclotomic polynomials.
Now note that
$$V_{n,p} := \prod_{x \in F_{n,p}^{*}} \phi_{n,p}(x)$$ is of the form
$$\sum_{j=0}^{p-1}A_{j,n,p}e^{\frac{2\pi i j}{p}}$$
and by symmetry
$$A_{0,n,p}=A_{1,n,p}...=A_{p-1,n,p} = A_{n,p}.$$
thus $V_{n,p} = 0$. Thus some $x^{*} \in F_{n,p}^{*}$ which satisfies $\phi_{n,p}(x^{*}) = 0$ is the configuration we are after.
A: You can easily do this recursively / by induction:
The sum of the first $5n$ positive integers is equal to
$$
S:=\sum_{j=1}^{5n}j = \frac{(5n)^2 + 5n}{2}.
$$
So every set has to sum to $\frac{S}{5} = \frac{5n^2+n}{2}$ which for the case that $n=2k+1$ (so $n$ is odd) comes out to $$s_k := 10k^2+11k+3.$$ Now assume you have some partition $B_1,...,B_5$ of $\{1,...,n\}$ for $k \in \Bbb N$ such that $\sum B_j = s_k$ holds. We'll now construct a partition for $k+1$ from this. First note that $s_{k+1} - s_k = 20k+21$, so we have to increase the sum of each set in our partition by $20k+21$. To do this we want to use the newly added $10$ numbers $5n + 1, ..., 5n+10 = 10k+6, ..., 10k+15$ which we may also write as $$a_j := 10k + (j+5), j=1,...,10.$$ Note that by summing the consecutive pairs from the front and back we get $$a_{10-(j-1)} + a_j = 10k + ((10-(j-1))+5) + 10k + (j+5) = 20k + 21, j=1,...,5$$ which is precisely the difference we desire. Also note that we have precisely $5$ such pairs, so the sets $$A_j = B_j \cup \{a_j, a_{11-j}\}, j=1,...,5$$ are a partition of the next odd number $n+2$.
By showing that there is such a partition for $k=1$ (so $n=3$) we've thus proven that there exists such a partition for every odd integer. This base case can be done by a bit of trial and error or simply writing a 5-line program to do the search for us. One example is $\{ \{ 10, 14\},\{1, 11, 12\}, \{9, 15\}, \{8, 3, 13\}, \{2, 4, 5, 6, 7\}\}$.
