Upper Bound for the Difference of Even Probability and Odd Probability in Hypergeometric Distribution

Let $X$ be a random variable following the hypergeometric distribution with parameters $N,K,n$, where $$Pr(X=k) = \frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}.$$ To make things easier, we bring symmetry and avoid the boundry issues by letting $K=\frac{N}{2}$ and $n<K$. For an odd $n$, we know by symmetry $$Pr(X~even)=Pr(X~odd).$$ For an even $n$, my conjecture is that $$\vert Pr(X~even)-Pr(X~odd) \vert \leq \frac{\binom{N}{n/2}}{\binom{N}{n}}.$$ I ran some simulations and the result shows that it should be true, but I have not been able to prove it so far. Could someone help me with that?

If this question is answered, a subsequent question is the case where $K$ is roughly $\frac{N}{2}$ but not exact. A corresponding condition is $\vert K-\frac{N}{2}\vert \leq c N$, for a constant $0<c<0.5$. We still assume $n<min (K, N-K)$ to avoid boundary issues. The conjecture is still the same in this case.

I try not to be greedy and will be much satisfied if only the first question is answered. Many thanks in advance!

In the case $N=2K$ the sum can be evaluated in closed form using generating functions, and satisfies the desired inequality.
In general, multiply by ${N \choose n} t^k z^n$ and sum over both $k \leq K$ and $n \leq N$ to get $$G_{N,K}(t,z) = (1+tz)^K (1+z)^{N-K}$$ (by writing $n=k+m$ and factoring into a product of two binomial sums). The desired alternating sum is then ${N \choose n}^{-1}$ times the $z^n$ coefficient of $G_{N,K}(-1,z) = (1-z)^K (1+z)^{N-K}$. When $K=N-K$, this simplifies to $(1-z^2)^K$, so the coefficient is zero for $n$ odd (as you noted by an antisymmetry argument) and $(-1)^n {K \choose n/2}$ for $n$ even. In particular the absolute value is less than $N \choose n/2$ as desired (and by a factor of about $2^{n/2}$). For general $K,N$, you can bound or estimate the $z^n$ coefficient using stationary-phase techniques.