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
$1\cdot 1mod15=1$
$2\cdot 8=16mod15=1$
$4\cdot 4=16mod15=1$
$7\cdot 13=91mod15=1$
$11\cdot 11=121mod15=1$
$14\cdot 14=196mod15=1$

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 $3\le n\le 2\times {10}^7$.