How many edge-disjoint copies of $s$-complete graph $K_s$ in the complete $s$-partite graph $K_{l,l,\dots,l}$ with each part of size $l$? How many edge-disjoint copies of $s$-complete graph $K_s$ in the complete $s$-partite graph $K_{l,l,\dots,l}$ with each part of size $l$?
For vertex-disjoint case, the answer is simple, just $l$ copies of $K_s$. But it seems there are much more edge-disjoint copies of $K_s$ in $K_{l,l,\dots,l}$. How many could it be? Is there any known result about it?
 A: We’ll follow wonce’s comment and denote the maximum size of a required family of $s$-tuples by $t(l,s)$. 
Clearly, if $s=1$ then we can take all $l$ $s$-tuples, so $t(l, 1)=l$. Farther we’ll assume that $l\ge 2$. 
Since there are at most $l^2$ distinct possibilities for the pair $(x_1,x_2)$, $t(l,s)\le l^2$.  
For a prime number $p$ we can show $t(p,p)\ge p^2$ by finding required $p^2$ $p$-tuples $t^{ij}$ in the product $\Bbb Z_p^p$  of the groups $\Bbb Z_p=\Bbb Z/p\Bbb Z$ of residues modulo $p$. For any $(i,j)\in \Bbb Z_p^2$ and $1\le k\le p$ put $t^{ij}_k\equiv i+j(k-1)\pmod p$. Assume that two tuples $t^{ij}$ and $t^{i’j’}$ have common $k_1$-th and $k_2$ elements. Then 
$i+j(k_1-1) \equiv i’+j’(k_1-1) \pmod p$ and $i+j(k_2-1)\equiv i’+j’(k_2-1) \pmod p$. Subtracting one of the equalities from the other, we obtain $(j-j’)(k_2-k_1)\equiv 0\pmod p$. Since $l$ is prime and $k_1\not\equiv k_2\pmod p$, we have $j\equiv j’\pmod p$, which implies $i\equiv i’\pmod p$.
Due to lower approximation of naturals by primes,
the above construction suggests that if $r=\min\{s,t\}$ then $t(l,s)\ge r^2-O(r\sqrt{r})$.
Now we consider small $l$. 
We have $t(2,3)=4$ as shows the family $\{(1,1,1), (1,2,2), (2,1,2), (2,2,1)\}$ of triples. I can show that $t(2,4)=2$, and then by induction I can prove that $t(l, (l^2+l)/2+1)=l$.
