Let $X_m \sim \mathcal {HG} (N,m,n)$, for which the expectation is given by $\mathbb E(X_m)=\frac{nm}{N}$. Assume that $N$ and $n$ are fixed, and we want to determine $m \in [N-1]$ that minimizes the cumulative distribution function (CDF) of $X_m$ at $\mathbb E(X_m)$, that is,
$$ \min_{m \in[N-1]} \mathbb P(X_m \le \mathbb E(X_m))= \frac { \sum_{0 \le i \le \frac {nm}{N} } \binom{m}{i} \binom{N-m}{n-i}}{\binom{N}{n}}.$$
Let $m^*$ denote an optimal solution of the above problem. From my numerical experiments in a part of another question [1], I realized that the following always holds:
Claim: If $n|N$ or $\frac nN \le \frac 14$, then $m^*=N-1$ and $\mathbb P(X_{m^*} \le \mathbb E(X_{m^*}))=\frac{n}{N}$.
I am looking for a proof or counterexample for the above claim.
Unfortunately, the quantity $\mathbb P(X_m \le \mathbb E(X_m))$ varies strangely with respect to $m$ and may have several local minima and maxima. Below, you can see four instances. The condition of the above claim is only satisfied for $N=99,n=33$ and $N=100, n=7$.
It is worth noting that by the dominance relation $X_{m+1}\succeq X_{m}$ [2], $\mathbb P(X_m \le x)$ is decreasing in $m$ for any fixed $x$. Above, $x$ is not fixed and given by $x=\frac{nm}{N}$, which depends on $m$.
Update 1:
I could find a way to prove the first part (you may see may answer).
Update 2:
I could present a proof for the second part using another method, which gives a stronger result:
If $\frac nN\le \frac 8{27}$, then $m^*=N-1$ and $\mathbb P(X_{m^*} \le \mathbb E(X_{m^*}))=\frac{n}{N}$.