Let the numbers excluded be

x - 1, x

and

x + 1

. Then we know that

\sum_{i = 1}^{n} i - 3 x = 7.5 \times (n - 3) (1) .

Hence:

[\sum_{i = 4}^{n} i] / (n - 3) \geq 7.5 \geq [\sum_{i = 1}^{n - 3} i] / (n - 3) .

It follows that:

n (n + 1) / 2 - 6 \geq 7.5 (n - 3) (2)

and

7.5 \geq \frac{1}{2} (n - 2) (3) .

Hence

n^{2} - 14 n - 33 = (n - 3) (n - 11) \geq 0

and, as we already know

n

is not less than 3, this gives

n \geq 11

From (2) and (3) we have

11 \leq n \leq 17 .

From (1): $n (n + 1) / 2 - 3 x = 7.5 (n - 3)$ and so

$x = [n (n + 1) / 2 - 7.5 (n - 3)] / 3$

Using a spreadsheet to calculate $x$ for $n \in [11, 17]$ we find that the solutions are:
$n = 11$ , remove 1,2, and 3
$n = 15$ , remove 9, 10 and 11
$n = 17$ , remove 15, 16 and 17.

See the spreadsheet giving the solutions.