Clearly, for $n=2014$ we can just take $a_{2k}=2014, a_{2k+1}=2015$ s.t. $P(-1)=0$. Assume we have such a polynomial with degree $2n<2 \cdot 2014$ which has a real root. Clearly $P(x) \ge 2014>0$ for $x \ge 0$ and hence the root must be negative. So $Q(x)=P(-x)$ must have a positive real root. But for any positive $x$ we have $Q(x) \ge 2014x^{2n}-2015x^{2n-1}+2014x^{2n-2}-\dotsc-2015x+2014=R(x)$. It is now sufficient to prove that $R(x)>0$ for $x>0$. We have $R(1)=2014(n+1)-2015n=2014-n>0$. So now assume $x \ne 1$. Then $R(x)>0$ is equivalent to $2014(x^{2n}+x^{2n-2}+\dotsc+1) \ge 2015(x^{2n-1}+\dotsc+x)$ which is likewise equivalent to $2014\frac{x^{2n+2}-1}{x^2-1} \ge 2015x\frac{x^{2n}-1}{x^2-1}$ equivalent to $\frac{x^{2n+2}-1}{x(x^{2n}-1)} \ge 1+\frac{1}{2014}$. Since $n<2014$ it is sufficient to prove $\frac{x^{2n+2}-1}{x(x^{2n}-1)} \ge 1+\frac{1}{n}=\frac{n+1}{n}$ which is equivalent (after the same transformations as above) to $nx^{2n}+nx^{2n-2}+\dotsc+nx^2+n \ge (n+1)x^{2n-1}+(n+1)x^{2n-3}+\dotsc+(n+1)x$ for all positive $x$. But for any $k$ we have $nx^{2k+1} \le \frac{n}{2}x^{2k}+\frac{n}{2}x^{2k+2}$ by AM-GM. Additionally we have $x^k+x^{2n-k} \le x^{2n}+1$ (which is equivalent to $(x^i-1)(x^{2n-i}-1) \ge 0$ which is clearly true). Adding the first type of inequality for all $k$ from $0$ to $n-1$ and the second type for all odd $k$ from $1$ to $2n-1$ we get the desired result. Hence for $n<2014$ such a polynomial can't exist. And thus, the answer is $\boxed{2014}$.

I just noticed that it is actually very similar to this Problem from the Book (19.3).