For which $n\in\mathbb{Z}^+$ is there a $\sigma\in S_{14}$ such that $|\sigma|=n$? For which $n\in\mathbb{Z}^+$ is there a $\sigma\in S_{14}$ (the group of permutations on $\{1,2,\dots,14\}$) such that $|\sigma|=n$ (where $|\sigma|$ is the order of $\sigma$)?
I know you could just enumerate all elements of $S_{14}$, but that would take a very long time. Are there any more elegant approaches?

The order of a cycle is the length of the cycle, and every permutation is a product of commuting cycles. Thus the possible orders are all least common multiples of sets of positive integers whose sums are less than or equal to 14.
By hand I found 29 distinct combinations yielding 13, 11, 22, 33, 9, 18, 36, 45, 8, 24, 40, 7, 14, 21, 28, 35, 42, 84, 5, 10, 15, 20, 30, 60, 4, 12, 3, 6, 2. The identity is also of order 1, so that makes  30 different orders.
EDIT: As Derek pointed out, I missed one. The culprit is cycle type $(7,5,2)$, which is of order 70.
