Mathematica competitions in Croatia

Arranging the expression we get $$\begin{array}{rl}\frac{n^8+n^6+n^4+n^2+a}{n^2-1}& =\frac{n^8-n^6+2(n^6-n^4)+3(n^4-n^2)+4(n^2-1)+a+4}{n^2-1}\\ & =n^6+2\cdot n^4+3\cdot n^2+4+\frac{a+4}{n^2-1}.\end{array}$$ Since $2014\equiv 1(\mathrm{mod}3)$, every power of $2014$ is congruent to $1$ modulo $3$. Therefore, $2014^6+2\cdot 2014^4+3\cdot 2014^2+4\equiv 1+2+3+4\equiv 1(\mathrm{mod}3)$. 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 $a=2\cdot 2014^2-6$.