SOLVED

Let $1=d_1<\cdots <d_{\tau(n)}=n$ be the divisors of $n$ and
\[G(n) = \sum_{1\leq i<\tau(n)}\frac{d_i}{d_{i+1}}.\]
Is it true that $G(n)\to \infty$ for almost all $n$? Can one prove an asymptotic formula for $\sum_{n\leq X}G(n)$?

Erdős writes it is 'easy' to prove $\frac{1}{X}\sum_{n\leq X}G(n)\to \infty$.

Terence Tao has observed that, for any divisor $m\mid n$, \[\frac{\tau(n/m)}{m} \leq G(n) \leq \tau(n),\] and hence for example $\tau(n)/4\leq G(n)\leq \tau(n)$ for even $n$. It is easy to then see that $G(n)$ grows on average, and in general behaves very similarly to $\tau(n)$ (and in particular the answer to the first question is yes). Tao suggests that this was a mistaken conjecture of Erdős, which he soon corrected a year later to [448].