No, it is not possible. In particular, consider that, were we to number the stakes from $0$ through $2015$ going evenly around the circle, then a string going from $a$ to $b$ is parallel to one going from $c$ to $d$ if and only if $a+b\equiv c+d\pmod{2016}$. This may be seen as there is a reflection taking stake $a$ to stake $b$ which also takes stake $c$ to stake $d$. Two sets of parallel lines between such labeling are shown below:
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/invnk.png)
Thus, we may reduce the problem to the following:
Is there a sequence $s_0,\ldots,s_{2015}$ such that the value of $s_n+s_{n+1}$ is distinct for every $n$ between $0$ and $2015$ (inclusively, taking the last value to be $s_{2015}+s_0$)?
And the answer is "no" because there are only $2016$ equivalence classes mod $2016$, meaning each equivalence class must contain exactly one value $s_{n}+s_{n+1}$. However, that means that $$1008\equiv\sum_{i=0}^{2015}i\equiv \sum_{n=0}^{2015}s_n+s_{n+1}\equiv 2\sum_{n=0}^{2015}s_n\equiv 2\sum_{i=0}^{2015}i\equiv2\cdot 1008\pmod{2016}$$
which is false since $2\cdot 1008 \not\equiv 1008$. Note that we used, in our second to last step, that $s_n$ enumerates the equivalence classes mod $2016$.
(Note that the equivalence $1008\equiv\sum_{i=0}^{2015}i\pmod{2016}$ comes from a pairing argument - each summand $i$ may be paired with its opposite $-i$ making a total of $0$, with the exception of the two $i$ such that $i=-i$ - which are $0$ and $1008$. So the sum is $0+1008=1008$)