3. If $a, b$ are two positive numbers, and $\frac{a-1}{b}+\frac{b-1}{a}+1=0$, then ( ).
(A) $0<a+b \leqslant \frac{1}{3}$
(B) $\frac{1}{3}<a+b \leqslant 1$
(C) $1<a+b \leqslant \frac{4}{3}$
(D) $\frac{4}{3}<a+b \leqslant 2$
3. C.

From the problem, we have $a^{2}+a b+b^{2}=a+b$.
Then $a b=(a+b)^{2}-(a+b)$
$$
=(a+b)(a+b-1) \text {. }
$$

Since $a, b$ are two positive numbers, then $a b>0, a+b>0$.
Therefore, $a+b-1>0 \Rightarrow a+b>1$.
By $a b \leqslant \frac{(a+b)^{2}}{4}$, and combining with equation (1), we get
$$
\frac{a+b}{4} \geqslant a+b-1 \Rightarrow a+b \leqslant \frac{4}{3} \text {. }
$$

Thus, $1<a+b \leqslant \frac{4}{3}$.