jmc

Let $x=2005$. Then the expression becomes $$(x+4{)}^4-4(x+2{)}^4+6x^4-4(x-2{)}^4+(x-4{)}^4$$We use the Binomial theorem (or Pascal's triangle) to expand the expression and get $$\begin{array}{rl}& x^4+4x^3\cdot 4+6x^2{4}^2+4x\cdot 4^3+4^4\\ & -4(x^4+4x^3\cdot 2+6x^2\cdot 2^2+4x\cdot 2^3+2^4)\\ & +6x^4\\ & -4(x^4+4x^3\cdot 2+6x^2\cdot 2^2+4x\cdot 2^3+2^4)\\ & +x^4+4x^3\cdot 4+6x^2{4}^2+4x\cdot 4^3+4^4.\end{array}$$Many of these terms cancel out! After simplifying, we are left with $$4^4-4\cdot 2^4-4\cdot 2^4+4^4=2\cdot 4^3(4-1)=128\cdot 3=\boxed{384}$$