Obviously not for every set $A \subset \mathbb{N}$ there is a group $G$ with $A$ as set of orders of its elements (usually called 'spectrum') -- for example if $G$ has an element of order $n$, then $G$ also has an element of order $d$ for every divisor $d$ of $n$.
For a survey of what is known on this question, you may check the following references: http://link.springer.com/article/10.1023%2FA%3A1014658001689?LI=true, http://mat.polsl.pl/groups/docs/Vasilev.pdf, http://www.math.nsc.ru/conference/malmeet/12/plenary/mazurov.pdf.