$\psi$ is the multiplicative convolution of $\mu^2$ and the identity function, hence its Dirichlet series is 
$$\sum_{n=1}^\infty\frac{\psi(n)}{n^s}=\frac{\zeta(s)\zeta(s-1)}{\zeta(2s)},\qquad\Re(s)>2.$$ 
This implies by [Perron's formula][1] and standard bounds that 
$$\sum_{n \leq X} \psi(n)\sim\frac{\zeta(2)}{2\zeta(4)}X^2=\frac{15}{2\pi^2}X^2.$$


  [1]: https://en.wikipedia.org/wiki/Perron%27s_formula