1. Given that $a$ and $b$ are non-negative real numbers, and
$$
a^{2005}+b^{2005}=1, a^{2006}+b^{2006}=1 \text {. }
$$

then $a^{2007}+b^{2007}=$ $\qquad$
Ni, 1.1.
If $a$ or $b$ is greater than 1, then $a^{2005} + b^{2005} > 1$; if $0 < a, b < 1$, then $a^{2005} > a^{2006}$, $b^{2005} > b^{2006}$. Therefore, $a^{2000} + b^{2000} > a^{2006} + b^{2000}$, which means $1 > 1$, a contradiction. Hence, one of $a$ or $b$ is 0, and the other is 1. Thus, $a^{200} + b^{20 n} = 1$.