In general the converse to Lagrange's Theorem is false - $A_4$ has no subgroup of order $6$.
However, the converse holds for the set of primes - given a prime $p$ and a finite group $G$, if $p$ divides $|G|$, then there exists a subgroup of $G$ of order $p$. This is Cauchy's Theorem.
Indeed, this also holds for prime powers $p^k$.
Are the prime powers the maximal set for which the converse of Lagrange's Theorem holds? In other words, does there exists some composite integer $n>1$, not a prime power, such that for any finite group $G$, if $n$ divides $|G|$ then $G$ has a subgroup of order $n$? Or is it the case that for any such $n$ one can construct a finite group $G$ with order divisible by $n$ yet with no subgroup of order $n$?
