Solution: By Cauchy-Schwarz inequality, we have
$$\begin{array}{l} \left(\sum_{i=1}^{n} a_{i}+\sum_{i=1}^{n} b_{i}\right)\left(\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}}\right) \\ \geqslant\left[\sum_{i=1}^{n} \frac{a_{i}}{\sqrt{a_{i}+b_{i}}} \sqrt{a_{i}+b_{i}}\right]^{2} \\ =\left(\sum_{i=1}^{n} a_{i}\right)^{2}=1 . \\ \text { Then } \sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}} \geqslant \frac{1}{2} . \end{array}$$
$$\text { When } a_{1}=a_{2}=\cdots=a_{n}=b_{1}=\cdots=b_{n}=\frac{1}{n}$$

the equality holds, at this time, $\sum_{i=1}^{n} \frac{a_{i}^{2}}{a_{i}+b_{i}}$ takes the minimum value $\frac{1}{2}$.