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$\psi$ 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 and standard bounds that $$\sum_{n \leq X} \psi(n)\sim\frac{\zeta(2)}{2\zeta(4)}X^2=\frac{15}{2\pi^2}X^2.$$