This question is motivated by an old math contest problem, and is a generalization of the original problem. I will write out the original problem as motivation.

Let us say that $n$ is $p$-Savage, for a prime $p$, if it is possible to partition $\{1, \cdots, n\}$ into $p$ sets $A_1, \cdots, A_{p}$ such that the following holds:

i) $\displaystyle \sum_{x \in A_i} x = D$, a constant, for $1 \leq i \leq p$;

ii) $x \in A_i$ implies that $x \equiv i-1 \pmod{p-1}$ for $1 \leq i \leq p-1$; and

iii) $x \equiv 0 \pmod{p}, 1 \leq x \leq n$ imply that $x \in A_p$.

Do there exist $p$-Savage numbers for every prime $p$? If so, can one estimate how many $p$-savage number there are say in $[1, N]$?

This is motivated by an old Canadian math contest problem (Euclid 2003), where the question asks about 3-Savage numbers following the above definition (originally just called savage numbers). It asked in particular to show that 8 is (3-)savage, and that every even savage number $n$ satisfies $n \equiv 8 \pmod{12}$. These questions are fairly simply to answer by doing some simple (but tedious) modular arithmetic. The contest did not ask for any sufficient conditions for $n$ to be savage, nor did it ask for whether the exist infinitely many savage numbers. It also only focused on even savage numbers. One can show by modular arithmetic that if $n$ is an odd savage number, then $n \equiv 11 \pmod{12}$. 35 is an odd savage number.