If $a^{2005}+b^{2005}=P(a+b,ab)$ then the requested sum equals $P(1,1)$. Therefore, take $a=\frac{1+i\sqrt{3}}{2}$ and $b=\frac{1-i\sqrt{3}}{2}$. We obtain $P(1,1)=1.$

I think have a solution (don't use complex number)

There is an official solution which use the period of g(x)=coeff sum of a^x+b^x expressed as the polynomial of a+b,ab , but i think it use the idea of complex number, which can produce odd period function. (i used complex number in the competition.i think it should be accepted.)

It is easy solved by induction. Let $a^n+b^n=P_n(x,y),x=a+b,y=ab.$ We have $P_1(x,y)=x,P_2(x,y)=x^2-2y,...$ $P_{n+1}(x,y)=xP_n(x,y)-yP_{n-1}.$ It give expression for polinom \[ P_n(x,y)=(\frac{x+\sqrt{x^2-4y}}{2})^n+(\frac{x-\sqrt{x^2-4y}}{2})^n=\frac{1}{2^{n-1}}\sum_{k=0}^{[n/2]} C_n^{2k}x^{n-2k}(x^2-4y)^k. \]

I only find the coefficients' sum of this polynomial$P(x)$ which $P(x+\frac{1}{x})=x^{2005}+\frac{1}{x^{2005}}$.This Polynomial is unique. defined by $P_{0}(x)=2,P_{1}(x)=x,P_{n+2}(x)=xP_{n+1}(x)-P_{n}(x)$