Hong Kong Preliminary Selection Contest

Let $\omega$ be a root of $x^2 + x + 1 = 0$. Then $\omega^3 = 1$ and $\omega \ne 1$.

Since $x^2 + x + 1$ divides $x^{2n} + x^n + 1$, we have $\omega^{2n} + \omega^n + 1 = 0$.

Let $y = \omega^n$. Then $y^2 + y + 1 = 0$, so $y = \omega$ or $y = \omega^2$.

Thus, $\omega^n = \omega$ or $\omega^n = \omega^2$.

This means $n \equiv 1 \pmod{3}$ or $n \equiv 2 \pmod{3}$.

So $n$ is any positive integer not exceeding $2014$ such that $n \not\equiv 0 \pmod{3}$.

The possible values of $n$ are those with $1 \le n \le 2014$ and $n \not\equiv 0 \pmod{3}$.

Let us compute the sum of all such $n$.

First, the sum of all $n$ from $1$ to $2014$ is:
$$S = 1 + 2 + \cdots + 2014 = \frac{2014 \times 2015}{2} = 2,029,105$$

Now, subtract the sum of all $n$ divisible by $3$ in this range.

The smallest such $n$ is $3$, the largest is $2013$.

The sequence is $3, 6, 9, \ldots, 2013$.

Number of terms:
Let $k$ be the number of terms. $3k = 2013 \implies k = 671$.

Sum of these terms:
$$S_3 = 3 + 6 + 9 + \cdots + 2013 = 3(1 + 2 + \cdots + 671) = 3 \times \frac{671 \times 672}{2} = 3 \times 225,456 = 676,368$$

Therefore, the sum of all $n$ not divisible by $3$ is:
$$2,029,105 - 676,368 = 1,352,737$$

Answer: $1,352,737$