, 2014

We have $n=2014$. First, for a special choice of the coefficients $a_{k}$, namely 2014 for even $k$ and 2015 for odd $k$, define the polynomial $Q_{n}(X)=2014 X^{2 n}+2015 X^{2 n-1}+2014 X^{2 n-2}+\ldots+2015 X+2014$. We have $Q_{2014}(-1)=0$, so the number sought is at most 2014. It remains to show that $P(X)$ has no real zero for $n \leq 2013$. For $x \geq 0$ we have $P(x) \geq a_{0}>0$, since the coefficients are positive; for $x<0$ we have $P(x) \geq Q_{n}(x)$, since $x^{k}$ is positive for even $k$, hence $a_{k} x^{k} \geq 2014 x^{k}$, and $x^{k}$ is negative for odd $k$, hence $a_{k} x^{k} \geq 2015 x^{k}$. It remains to show that $Q_{m}(x)>0$ for $m \leq 2013$.

1st proof: One computes that for real $x$ we have
$$Q_{m}(x)=1007\left(\sum_{\nu=0}^{m-1}\left(x^{\nu}+x^{\nu+1}\right)^{2}\right)+\frac{1}{2}\left(\sum_{\nu=0}^{m-1}\left(x^{2 \nu+1}+1\right)\left(x^{2 m-2 \nu-1}+1\right)\right)+\left(1007-\frac{m}{2}\right)\left(x^{2 m}+1\right)$$
The summands of the first sum are non-negative, being squares. Since $x^{2 k+1}+1$ is positive, 0, or negative, respectively, for all $k \geq 0$ for $x>-1$, $x=-1$, or $x<-1$, respectively, the summands in the second sum are non-negative. Thus for $x<0$ and $m \leq 2013$ we have:
$$P(x) \geq Q_{m}(x) \geq\left(1007-\frac{m}{2}\right)\left(x^{2 m}+1\right)>\frac{1}{2}$$

2nd proof (using analysis): For $x<0$ define the real function $f_{m}(x)=\left(x^{2}-1\right) Q_{m}(x)=2014\left(x^{2 m+2}-1\right)+2015 x\left(x^{2 m}-1\right)$. The second derivative $f_{m}^{\prime \prime}(x)=2(2 m+1)\left(2014(m+1) x^{2 m}+2015 m x^{2 m-1}\right)$ has the unique negative zero $x_{m}=-\frac{2015 m}{2014(m+1)}>-1$ for $m \leq 2013$, and we have
$$\begin{aligned} f_{m}^{\prime}\left(x_{m}\right) & =2014(2 m+2) x_{m}^{2 m+1}+2015(2 m+1) x_{m}^{2 m}-2015= \\ & =\left(2014(2 m+2) x_{m}+2015(2 m+1)\right) x_{m}^{2 m}-2015=2015 \cdot x_{m}^{2 m}-2015<0 \end{aligned}$$
For $x<x_{m}$ or $x_{m}<x<0$, respectively, $f_{m}^{\prime}$ is strictly increasing or strictly decreasing, respectively, since the derivative $f_{m}^{\prime \prime}$ is positive or negative, respectively; hence $f_{m}^{\prime}$ has a global maximum at $x=x_{m}$ for $x<0$ and is negative for all $x<0$. Hence $f_{m}$ is strictly decreasing, and thus, because of $f_{m}(-1)=0$, is positive for $x>-1$ and negative for $x<-1$. Thus $Q_{m}(x)$ is positive for $x \neq-1$; for $x=-1$ we directly get $Q_{m}(-1)=2014-m>0$.