Note that abs (n) = sqrt (n^2)
And abs ( n -3) = sqrt [ (n -3)^2 ]
And 9 = sqrt (9^2)
So
sqrt (n^2) < sqrt [ (n -3)^2 ]
n^2 < (n - 3)^2
n^2 < n^2 - 6n + 9
6n < 9
n < 9/6
n < 3/2
And
sqrt [ (n -3)^2 ] < sqrt (9^2)
(n - 3)^2 < 9^2
n^2 - 6n + 9 < 81
n^2 - 6n - 72 < 0
(n - 12) ( n + 6) < 0
This is true if n is on the interval
( -6 , 12)
Choosing the least restrictive interval we have that
n = (-6, 3/2)
The integers in this interval = -5 , -4 ,-3, -2 , -1, 0 , 1
Their sum = -14