How Many Even Coefficients?

Let $g(x) = \displaystyle \sum_{i=0}^{2014/2} a_i x_i$ and $h(x) = \displaystyle \sum_{i=0}^{2014/2} b_i x^i.$ We claim that in order for $f(x)$ to have all its coefficients even, either all coefficients of $g(x)$ must be even or all coefficients of $h(x)$ must be even. Assume the contrary. Let $p$ be the smallest index for which $a_p$ is odd, and let $q$ be the smallest index for which $b_q$ is odd. Note that the coefficient of $x^{p+q}$ in $f(x)$ is $\displaystyle \sum_{i=0}^{p+q} a_i b_{p+q-i}.$ The only odd term in this summation is $a_p b_q,$ since $a_i$ is even for all $i<p$ and $b_i$ is even for all $i<q.$ Thus, the coefficient of $x^{p+q}$ in $f(x)$ is odd, contradiction.

It follows that atleast one of $N_g$ or $N_h$ must be $\dfrac{2014}{2}+1,$ so $N_g+N_h \geq 1008.$ We can set all coefficients of $g(x)$ even and all coefficients of $h(x)$ odd. Note that this satisfies the problem condition, since if $a_i = 2c_i$ for all $i$ for some $c_i \in \mathbb{N},$ $f(x) = g(x) h(x) = 2 \left( \displaystyle \sum_{i=0}^{2014/2} c_i x^i \right) \left( \displaystyle \sum_{i=0}^{2014/2} b_i x_i\right),$ whose coefficients are clearly even. Thus, $N_g+N_h= 1008$ is attainable, and our answer is $1008-1= \boxed{1007}.$