The answers to this problem are:

either $a = 3/2$, $b = 2/3$, $c = 3$, $d = 1$ and $e = 4$

or $a = -3/2$, $b = -2/3$, $c = -3$, $d = -1$ and $e = -4$

*Roderick (Simon Langton Grammar School)
provided a correct solution based on an interesting trial and
improvement technique. Students at Smithdon High School - Robert ,
Jack and Matthew and John of Hethersett High School relied on a
more traditional method of proving their solution to the problem,
the essence of which is reproduced below.*

Let $a = N$

hence $b = 1/N$

and $c = 2/b$ i.e. $c = 2N$

and $d = 3/c$ i.e. $d = 3/2N$

and $e = 4/d$ i.e. $e = 4/(3/2n)$ i.e. $e = 8N/3$.

But $e = 6/N$.

Hence $6/N = 8N/3$

i.e $8N^2 = 18$

i.e $a = N = +$ or $- 3/2$ and so on....