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.

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:

H. Deng, M. S. Lucido, W. Shi: The Number of Isomorphism Classes of Finite Groups with Given Element Orders. Algebra and Logic 41 (2002), Issue 1, 39-46.

Andrey Vasil'ev: On finite groups with the given set of element orders. Talk slides, 2010.

V. D. Mazurov: Periodic groups with given element orders. Talk slides, Mal'tsev Meeting, Novosibirsk, November 12-16, 2012.