I am also unsure of what "nontriviality" conditions you want to impose. Without any further conditions, the following answers your question:
Call a positive integer $n$ nilpotent if every group of order $n$ is nilpotent.
Call a positive integer $n$ abelian if every group of order $n$ is abelian.
Suppose that the prime factorization of $n$ is $p_1^{a_1} cdots p_r^{a_r}$. Then:
1) $n$ is nilpotent iff for all $i,j,k$ with $1 leq k leq a_i$, $p_i^k not equiv 1 pmod{p_j}$.
2) $n$ is abelian iff it is nilpotent and $a_i leq 2$ for all $i$.
These results are proved in
Pakianathan, Jonathan(1-WI); Shankar, Krishnan(1-MI)
Nilpotent numbers.
Amer. Math. Monthly 107 (2000), no. 7, 631--634.
The proofs are constructive: for any $n$ which is not nilpotent (resp. abelian), they give an explicit group of that order which is not nilpotent (resp. abelian).
The paper is available at
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