Croatian Mathematical Competitions

Arranging the expression we get
$$\begin{aligned} \frac{n^8 + n^6 + n^4 + n^2 + a}{n^2 - 1} &= \frac{n^8 - n^6 + 2(n^6 - n^4) + 3(n^4 - n^2) + 4(n^2 - 1) + a + 4}{n^2 - 1} \\ &= n^6 + 2 \cdot n^4 + 3 \cdot n^2 + 4 + \frac{a+4}{n^2-1}. \end{aligned}$$
Since $2014 \equiv 1 \pmod{3}$, every power of $2014$ is congruent to $1$ modulo $3$.
Therefore, $2014^6 + 2 \cdot 2014^4 + 3 \cdot 2014^2 + 4 \equiv 1 + 2 + 3 + 4 \equiv 1 \pmod{3}$.
In order for the entire expression to be divisible by $3$, the number $\frac{a+4}{2014^2-1}$ has to be congruent to $2$ modulo $3$, so the least positive integer $a$ with the desired property is the one that satisfies $\frac{a+4}{2014^2-1} = 2$, which gives $a = 2 \cdot 2014^2 - 6$.