To determine when the function \(f(x) = x^3 - 3x^2 - 9x + 2\) is increasing, we need to find when its derivative \(f'(x)\) is non-negative.

First, find the derivative:

\(f'(x) = 3x^2 - 6x - 9\).

Set \(f'(x) \geq 0\) to find the critical points:

\(3x^2 - 6x - 9 = 0\).

Factor the quadratic equation:

\(3(x^2 - 2x - 3) = 0\).

\(3(x - 3)(x + 1) = 0\).

The critical points are \(x = 3\) and \(x = -1\).

Since \(f(x)\) is increasing for \(x > n\), we need \(f'(x) \geq 0\) for \(x > n\).

Test intervals around the critical points:

- For \(x > 3\), \(f'(x) > 0\).

- For \(-1 < x < 3\), \(f'(x) < 0\).

- For \(x < -1\), \(f'(x) > 0\).

Thus, \(f(x)\) is increasing for \(x > 3\).

Therefore, the least possible value of \(n\) is 3.