解:因为$(-2x^{m + 1}y^{2n - 1})\cdot5x^{n}y^{m}=-10x^{m + n + 1}y^{m + 2n - 1}=-10x^{4}y^{4},$
所以$\begin{cases}m + n + 1 = 4\\m + 2n - 1 = 4\end{cases},$
由$m + n + 1 = 4$可得$m=3 - n,$
把$m=3 - n$代入$m + 2n - 1 = 4,$得$3 - n + 2n - 1 = 4,$$n + 2 = 4,$解得$n = 2,$
把$n = 2$代入$m=3 - n,$得$m = 3 - 2 = 1。$
则$-2m^{2}n\cdot(-\frac{1}{2}m^{3}n^{2})^{2}=-2m^{2}n\cdot\frac{1}{4}m^{6}n^{4}=-\frac{1}{2}m^{8}n^{5},$
把$m = 1,$$n = 2$代入$-\frac{1}{2}m^{8}n^{5},$得$-\frac{1}{2}\times1^{8}\times2^{5}=-\frac{1}{2}\times32=-16。$