You Can't Do That With Wilson's Theorem!

Partial solution :

Note that $N \equiv 82!! \equiv 2^{41} \cdot 41! \pmod{83}$ and $83$ is prime, so in particular $\begin{aligned}N^2 &\equiv 2^{82} \cdot (41!)^2 \pmod{83}& \\ &\equiv 1 \cdot (41!)^2 \pmod{83} & \text{ by Fermat's Little Theorem} \\ &\equiv 41! \cdot (-1)^{41} (-1)(-2)\cdots (-41) \equiv 41! \cdot (-1) \cdot (82)(81)\cdots(42) \equiv (-1)\cdot 82! \pmod{83} & \\ &\equiv (-1) \cdot (-1) \pmod{83} & \text{ by Wilson's theorem} \\ &\equiv 1 \pmod{83} \end{aligned}$

Then using the fact $83$ is prime once more, $N^2 \equiv 1 \pmod{83} \Rightarrow N \equiv \pm 1 \pmod{83}$

Therefore the only two possible answers are $N=1$ or $N=82$ . I don't know yet how to show $N$ is not $1$ (other than brute force).