4. Given an odd function $f(x)$ defined on $\mathbf{R}$ whose graph is symmetric about the line $x=2$, when $0<x \leqslant 2$, $f(x)$ $=x+1$. Then $f(-100)+f(-101)=$ $\qquad$
4. 2 .

From the problem, we know that
$$
f(-x)=-f(x) \text {, and } f(2-x)=f(2+x) \text {. }
$$

Then, $f(x+4)=f(-x)=-f(x)$,
$$
f(x+8)=-f(x+4)=f(x) \text {. }
$$

Thus, $f(x)$ is a periodic function with a period of 8.
Also, $f(3)=f(1)=2, f(4)=f(0)=0$, so
$$
f(-100)+f(-101)=f(4)+f(3)=2 \text {. }
$$