Balls and Bins analysis in the $m \gg n$ regime: gaps
Answer: $\Theta\left(\sqrt{\frac{m}{n\log n}}\right)$.

Applying a multidimensional version of the Central Limit Theorem, we get that the vector $(X_1,\dots, X_n)$ has asymptotically multivariate Gaussian distribution with 
$$\mathrm{Var}[X_i] = m\left(\frac{1}{n} - \frac{1}{n^2}\right),$$
and
$$\mathrm{Cov}(X_i, X_j) = -m/n^2.$$
We will assume below that $X$ is a Gaussian vector (and not only approximately a Gaussian vector). Let us add a Gaussian random variable $Z$ with variance $m/n^2$ to all $X_i$ ($Z$ is independent from all $X_i$). That is, let
$$
\begin{pmatrix}
Y_1\\Y_2\\ \vdots\\Y_n 
\end{pmatrix}
=
\begin{pmatrix}
X_1+Z\\X_2+Z\\ \vdots\\X_n +Z
\end{pmatrix}.
$$
We get a Gaussian vector $(Y_1, \dots, Y_n)$. Now each $Y_i$ has variance $m/n$:
$$\mathrm{Var}[Y_i] = \mathrm{Var}[X_i] + \underbrace{2\mathrm{Cov}(X_i,Z)}_{=\, 0}+\mathrm{Var}[Z] = m/n,$$
and all $Y_i$ are independent:
$$\mathrm{Cov}(Y_i, Y_j) = \mathrm{Cov}(X_i, X_j) + \underbrace{\mathrm{Cov}(X_i,Z) + \mathrm{Cov}(X_j,Z)}_{=\, 0} +\mathrm{Cov}(Z, Z) = 0.$$

Note that $Y_i - Y_j = X_i - X_j$. Thus our original problem is equivalent to the problem of finding $Y_{\mathrm{max}} - Y_{\mathrm{sec-max}}$. Let us first for simplicity analyze the case when all $Y_i$ have variance $1$.


  Problem. We are given $n$ independent Gaussian r.v. $\gamma_1,\dots, \gamma_n$ with mean $\mu$ and variance $1$. Estimate the expectation of $\gamma_{\mathrm{max}} - \gamma_{\mathrm{sec-max}}$.


Answer: $\Theta\left(\frac{1}{\sqrt{\log n}}\right)$.

Informal Proof. 
Here is an informal solution to this problem (it's not hard to make it formal).
Since the answer does not depend on the mean, we assume that $\mu = 0$. Let $\bar\Phi(t) = \Pr[\gamma &gt; t]$, where $\gamma\sim{\cal N}(0,1)$. We have (for moderately large $t$),
$$\bar\Phi(t)\approx \frac{1}{\sqrt{2\pi}t} e^{-\frac{1}{2}t^2}.$$

Note that


$\Phi(\gamma_i)$ are uniformly and independently distributed on $[0,1]$, 
$\Phi(\gamma_{\mathrm{max}})$ is the smallest among $\Phi(\gamma_i)$,
$\Phi(\gamma_{\mathrm{sec-max})}$ is the second smallest among $\Phi(\gamma_i)$.


Thus $\Phi(\gamma_{\mathrm{max}})$ is close to $1/n$ and $\Phi(\gamma_{\mathrm{max}})$ is close to $2/n$ (there is no concentration but if we don't care about constants these estimates are good enough; in fact, they are even pretty good if we care about constants — but that needs a justification). Using the formula for $\bar\Phi(t)$, we get that
$$
2\approx \bar\Phi(\gamma_{\mathrm{sec-max}})\left/\bar\Phi(\gamma_{\mathrm{max}})\right. \approx e^{\frac{1}{2}\left(\gamma_{\mathrm{max}}^2 - \gamma_{\mathrm{sec-max}}^2\right)}.
$$

Thus $\gamma_{\mathrm{max}}^2 - \gamma_{\mathrm{sec-max}}^2$ is $\Theta(1)$ w.h.p. Note that $\gamma_{\mathrm{max}}\approx \gamma_{\mathrm{sec-max}} = \Theta(\sqrt{\log n})$. We have, 
$$\gamma_{\mathrm{max}} - \gamma_{\mathrm{sec-max}}\approx \frac{\Theta(1)}{\gamma_{\mathrm{max}} + \gamma_{\mathrm{sec-max}}} \approx \frac{\Theta(1)}{\sqrt{\log n}}.$$ 

QED

We get that 
\begin{align}
\mathbb{E}[{X_{\mathrm{max}} - X_{\mathrm{sec-max}}}] &amp;= \mathbb{E}[{Y_{\mathrm{max}} - Y_{\mathrm{sec-max}}}] \\
&amp;= \sqrt{\mathrm{Var}[Y_i]} \times\mathbb{E}[{\gamma_{\mathrm{max}} - \gamma_{\mathrm{sec-max}}}] = \Theta\left(\sqrt{\frac{m}{n\log n}}\right).
\end{align}


  The same argument goes through when we have arbitrary scores. It shows that 
  $$\mathbb{E}[X_{\mathrm{max}}- X_{\mathrm{sec-max}}] = c\, \left. \mathbb{E}[X_{\mathrm{max}}- X_{\mathrm{min}}]\right/\log n.$$