1. If $a, b$ are positive numbers, and
$$
a^{2009}+b^{2009}=a^{2007}+b^{2007} \text {, }
$$

then the maximum value of $a^{2}+b^{2}$ is $\qquad$
Given $a, b$, without loss of generality, assume $a \geqslant b>0$.
Then $a^{2} \geqslant b^{2}, a^{2007} \geqslant b^{2007}$.
$$
\begin{array}{l}
\text { At this point, }\left(a^{2007}-b^{2007}\right)\left(a^{2}-b^{2}\right) \geqslant 0 \\
\Rightarrow a^{2009}+b^{2009} \geqslant a^{2} b^{2007}+a^{2007} b^{2} \\
\Rightarrow\left(a^{2007}+b^{2007}\right)\left(a^{2}+b^{2}\right) \\
\quad=a^{2009}+b^{2009}+a^{2} b^{2007}+a^{2007} b^{2} \\
\quad \leqslant 2\left(a^{2009}+b^{2009}\right) .
\end{array}
$$

Considering $a^{2009}+b^{2009}=a^{2007}+b^{2007}$, and both sides are positive, so $a^{2}+b^{2} \leqslant 2$.
When $a=b=1$, $a^{2}+b^{2}=2$.