$$\text{Solve } \begin{aligned} \sum_{d \mid \varepsilon_{e}} \lambda(d) & =(-1)^{0}+(-1)^{1}+\cdots+(-1)^{\alpha} \\ & =\left\{\begin{array}{ll} 1, & 2 \mid \alpha, \\ 0, & 2 \nmid \alpha . \end{array}\right. \end{aligned}$$

From this and the fact that $\lambda(n)$ is a multiplicative function, we get
$$\sum_{d \mid n} \lambda(d)=\left\{\begin{array}{ll} 1, & n \text{ is a perfect square, } \\ 0, & \text{ otherwise. } \end{array}\right.$$