smc

Note that $$4000\cdot {\left(\frac{2}{5}\right)}^n=\left(2^5\cdot 5^3\right)\cdot {\left(2\cdot 5^{-1}\right)}^n=2^{5+n}\cdot 5^{3-n}.$$ Since this expression is an integer, we need: 1. $5+n\ge 0,$ from which $n\ge -5.$ 2. $3-n\ge 0,$ from which $n\le 3.$ Taking the intersection gives $-5\le n\le 3.$ So, there are $3-(-5)+1=\boxed{9}$ integer values of $n.$