Derived

By Maclaurin series , we have:

$\begin{aligned} x^2e^{x^2} & = x^2 \left(1 + x^2 + \frac {x^4}{2!} + \frac {x^6}{3!} + \cdots \right) \\ & = x^2 + x^4 + \frac {x^6}{2!} + \frac {x^8}{3!} + \cdots \\ \frac {d^2}{dx^2} x^2e^{x^2} & = 2! + (4\cdot 3)x^2 + \frac {(6 \cdot 5)x^4}{2!} + \frac {(8 \cdot 7)x^6}{3!} + \cdots \\ \frac {d^4}{dx^4} x^2e^{x^2} & = \frac {4!}{2!} + \frac {(6\cdot 5 \cdot 4 \cdot 3)x^2}{3!} + \frac {(8\cdot 7 \cdot 6 \cdot 5)x^4}{4!} + \cdots \\ \implies \frac {d^{2018}}{dx^{2018}} x^2e^{x^2} & = \frac {2018!}{1008!} + \frac {2020!x^2}{2! \cdot 1009!} + \frac {2022!x^4}{4! \cdot 1010!} + O(x^6) \\ \implies M & = \frac {d^{2018}}{dx^{2018}} x^2e^{x^2} \bigg|_{x=0} = \frac {2018!}{1008!} = 1009 \times 1010 \times 1011 \times \cdots 2018 \end{aligned}$

The smallest and largest 4-digit prime factors of $M$ are 1009 and 2017 respectively and their sum $1009+2017 = \boxed{3026}$ .