Dividing tournament into "equal" groups In a tournament of $n$ teams, each team plays all other teams exactly once, with no draw. For which $n$ is it always possible to divide all teams into several groups so that each group of teams won the same number of games in total? 
[Source: Ukrainian competition problem]
 A: I finally think (I believe :p) that I understood the question. The word "tournament" is really not appropriated and confuse because the description if of a league (match of everyone vs everyone), tournaments generally have structure of tree, what is not the case here.
For any number of points where we want connect every point with another by just a line this mean that we are choosing the ways to have 2 points, i.e. if the number of teams is n the number of matches will be 
$$m=\binom{n}{2}=\frac{n(n-1)}{2}$$
We can see that m victories can be distributed on n teams with some restrictions: the maximum number of victories some team can acquire is $n-1$ and JUST ONE team can do it because every team play exactly $n-1$ matches.
There is another restraint: only one team can lose all games, by the same reason that the maximum. So we have a possible and unique maximum of victories that is $n-1$ and a possible and unique 'pure lose', i.e. zero victories for some team.
Between $n-1$ and $0$ the m victories can be distributed in any way between teams.
STRATEGY
If I can create a distribution with and odd number of victories for some team and an even number of victories for all the others then it must be impossible to rearrange teams on groups with the same number of victories, because the sum of even numbers is even and the sum of an odd number and any amount of even numbers is odd.
A different strategy will be prove that exist some distribution of m with 2 or 3 kind of number of victories that make impossible rearrange them in groups with the same number of victories.
I need to prove that these kind of distributions exists for n positions and m victories with the restraints of one maximum $(n-1)$ and just one possible $0$.
I will assume that you can rearrange teams on groups of different cardinality.
FOR EVEN n
If n is even then I can prove that only $n=4$ can be rearranged on groups with same number of victories, assuming that you can create groups with different length. 
If n is even then $n-1$ is odd, so I can see if subtracting some odd quantity to m the rest can be divided on some a quantity:
$$a=\frac{m-(n-1)}{n-1}=\frac{\frac{n(n-1)}{2}-(n-1)}{n-1}=\\
a=\frac{n-2}{2}$$
I must see now if $n-1$ can be expressed as sums of $a$ quantities, i.e. if $$n-1=k\cdot a=k\frac{n-2}{2}\to n=2+\frac{2}{k-2};\ k\in\Bbb N$$ 
For $k=3\to n=4$ is the only case with even n where $n-1$ can be expressed as sums of $a$.
We can extend this analysis on this distribution $m_e=\{n-1,a,...,a\}$ to the cases where we want to see if is possible to express any sum $(n-1)+k\cdot a=j\cdot a;\ k,j\in\Bbb N$. We can see that this is impossible for any n except $n=0$ or $n=1$ what is out of this question because for a match we need, at least, 2 teams, i.e. $n>1$.
Conclusion: for any other even $n\neq 4$ will exist a distribution of the kind $m_e=\{n-1,a,a,...,a\}$ where is impossible distribute this m victories on groups with the same number of victories.
FOR ODD N
Now we can try something similar for odd n. I must try to prove that for every odd n exist a distribution of the kind $m_o=\{n-1,n-2,a,a,a,...,a\}$, where $n-2$ is an odd number on this distribution and the amount $(n-1)+(n-2)$ cannot be expressed as sums of $a$ (obviously $(n-1)$ and $(n-2)$ annot be expressed, at the same time, as multiple $a$).
I will show that this $a$ exists.
$$a=\frac{m-(n-1)-(n-2)}{n-2}=\frac{n^2-5n+6}{2n-4}\to n^2-n(5+2a)+6+4a=0\\
n=2a+3;\ a=\frac{n-3}{2}$$
Now I must see if the quantity $(n-1)+(n-2)$ can be expressed on terms of $a$ AND AT THE SAME TIME this quantity $(n-1)+(n-2)=k\cdot a$ can be a multiple of $n-2$ positions that have a value of $a$, i.e. $j\cdot a=n-2$, i.e.
$$2n-3=k\frac{n-3}{2}\to n=3+\frac{6}{k-4};\ k\in\Bbb N$$
and
$$j\cdot k\cdot a=n-2\to n=3+\frac{2}{jk-2};\ k,j\in\Bbb N$$
For the first formula we can see that $k\in\{1,2,3,5,6,7,10\}$ leading to $n\in\{1,0,-3,9,6,5,4\}$ respectively. The only interesting cases here are the odd n, i.e. $k\in\{3,5,7\}$. This means that for these k the quantity $(n-1)+(n-2)=k\cdot a$.
For every $k$ we see that the $j$ in the second formula lead to a contradiction or inexistent $j$. So doesnt exist a distribution of the kind $m_o=\{n-1,n-2,a,...,a\}$ that can express the quantity $(n-1)+(n-2)$ as sums of $k\cdot a$ and at the same time this $k\cdot a$ be proportional to the length $(n-2)$.
We can prove too that the quantity $(n-1)+(n-2)+a$ (remember $a=\frac{n-3}{2}$) cannot be expressed as a sum of $k\cdot a$ and at the same time $j\cdot k\cdot a=n-3$ for any $j$, i.e., be possible divide the distribution in this kind of sums. For bigger sums of the kind $(n-1)+(n-2)+h\cdot a$ you can prove the same (the unique possible cases of study are $n=5$ and $n=9$).
So this prove that for every odd $n$ exist a distribution of the kind $m_o=\{n-1,n-2,a,a,...,a\}$ that make impossible distribute these values on groups with the same value.
EDIT: I need a last proof on the distribution $m_o$, if it possible some distribution of the values $(n-1)+k\cdot a=(n-2)+j\cdot a=h\cdot a$ for some $k+j+l\cdot h=n-2;\ k,j,l\in\Bbb N$ and $h\in\Bbb N_0$.
We can start seeing if is possible that $(n-1)+ka=(n-2)+ja\to a=\frac{1}{j-k}$. This leave the only case with sense where $a=1\to n=5$. We can see that the distribution for $n=5,\ m_o=\{4,3,1,1,1\}$ can be arranged, effectively, as two groups of the same amount of victories, i.e. $m_o=\{4,1\}+\{3,1,1\}$.
But for other side we can see that the distribution for $n=5,\ m^*=\{4,0,2,2,2\}$ cannot be rearranged on groups with the same value in any way.
Conclusion: doesnt exist any odd n that can lead to any kind of distribution of it m victories that make possible rearrange them on groups with same number of victories.
CONCLUSION
So the unique n that can have it m victories distributed on groups of the same number of victories is $n=4$. 
For any other n will exist a distribution of the kind $m_e=\{n-1,a,...,a\}$ for even n, and $m_o=\{n-1,n-2,a,a,...,a\}$ for odd $n\neq 5$ that make impossible distribute the m victories in groups with the same number of victories. And for $n=5$ we have the distribution $m^*=\{4,0,2,2,2\}$ that cannot be grouped in any valid way.
P.S.: if I assume that we divide the m victories on groups of the same length then doesnt exist any n that can distribute the victories on the n teams to have groups with the same number of victories.
P.S.2: I dont know anything about graph theory so possibly my answer is unexpectedly too long. Sry for too much editions... a lot of tiny (some ones big) mistakes.
