1. Given that $f(x)$ is an odd function defined on $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have
$$
f(2+x)+f(2-x)=0 .
$$

If for $x \in[-1,0)$,
$$
f(x)=\log _{2}(1-x) \text {, }
$$

then $f(1)+f(2)+\cdots+f(2021)=$ $\qquad$
- 1. -1 .

From $f(x)$ being an odd function defined on $\mathbf{R}$, we get
$$
f(0)=0, f(1)=-f(-1)=-1 \text {. }
$$

Taking $x=0$, we get
$$
f(2)=-f(2) \Rightarrow f(2)=0 \text {. }
$$

Let $y=2-x$, then
$$
\begin{array}{c}
f(y)=-f(2+(2-y)) \\
=-f(4-y)=f(y-4) .
\end{array}
$$

Thus, $f(x)$ is a periodic function with a period of 4.
$$
\text { Therefore, } f(3)=f(-1)=1, f(4)=f(0)=0 \text {. }
$$

Hence, the required sum is
$$
505 \times(f(1)+f(2)+f(3)+f(4))+f(1)=-1 .
$$