Eccoci.

Consider the number $15$.
There are eight positive numbers less than $15$ which are coprime to $15$: $1, 2, 4, 7, 8, 11, 13, 14$.
The modular inverses of these numbers modulo $15$ are: $1, 8, 4, 13, 2, 11, 7, 14$
because
$1 \cdot 1 \bmod 15=1$
$2 \cdot 8=16 \bmod 15=1$
$4 \cdot 4=16 \bmod 15=1$
$7 \cdot 13=91 \bmod 15=1$
$11 \cdot 11=121 \bmod 15=1$
$14 \cdot 14=196 \bmod 15=1
$

Let $I(n)$ be the largest positive number $m$ smaller than $n-1$ such that the modular inverse of $m$ modulo $n$ equals $m$ itself.
So $I(15)=11$.
Also $I(100)=51$ and $I(7)=1$.

Find $\sum I(n)$ for $3 \le n \le 2 \times 10^7$.