Has this anything to do with number theory at all, other than that we need $ p|\frac {k(k + 1)}2$?

Does anyone knows the answer? What about the case p=2 for a start?

alphachapmtl wrote: Does anyone knows the answer? What about the case p=2 for a start? For p=2, we need $ k\equiv 0,3 \bmod{4}$, as noted by Peter, since we must have $ p\mid (1+2+...+k)$ If $ k\equiv 0 \bmod{4}$, we make the sets $ \{1,4,5,8,\cdots,i,i+3,\cdots,k-3,k\}$ and $ \{2,3,6,7,\cdots ,i+1,i+2,\cdots,k-2,k-1\}$, because $ i+i+3=i+1+i+2$. If $ k\equiv 3 \bmod{4}$, we make the sets $ \{1,2\}\cup\{4,7,\cdots ,i,i+3,\cdots,k-3,k\}$ and $ \{3\}\cup\{5,6,\cdots,i+1,i+2,\cdots,k-2,k-1\}$, which again work because $ i+i+3=i+1+i+2$.

Peter wrote: Has this anything to do with number theory at all, other than that we need $ p|\frac {k(k + 1)}2$? You're right,$ p|\frac {k(k + 1)}2$.Using simple analysis,we can reduce it to $ k=2p,3p,2p-1,3p-1$ Then dealing with these four cases is not difficult

If $ k$ is solution, $ k+2p$ - solution too. We can add $ S_i, i=1,2,...,p$ for k new elements $ \{S_i,k+i,k+2p-i\}$. If $ k=2p$, then $ S_i=\{i,2p+1-i\}$ - solution. If $ k=2p-1$, then $ S_i=\{i,2p-1-i\},i=1,..,p-1, S_p=\{2p-1\}$ is solution. Therefore $ k=np,np-1$ are solutions, when n is even. If $ p=2$ $ k=np,np-1$ are solutions for any n.

If $ p = 2$ then $ k = 4n - 1,4n$,not true for any $ n$.

how would one partition 3p-1, 3p?

for $3p$ : take $\{ i, \frac{3p+2-i}{2} , \frac{6p+1-i}{2} \} (1 \le i \le p , i \equiv 1 (mod 2) ) , \{ i, \frac{4p+2-i}{2} , \frac{5p+1-i}{2} \} (1 \le i \le p , i \equiv 0 (mod 2) )$ for $3p-1$ : take $\{ i , \frac{3p-1}{2}+i , 3p-2i-1 \} , \{ 3p-2i, \frac{p-1}{2}+i , p+i-1 \} (1 \le i \le \frac{p-1}{2}) $