Let $X_i$ be the coordinates of a centered $n$-variate normal distribution satisfying $$ \mathrm{cov}(X_i,X_j) = \begin{cases} 1 & \text{if }\; i=j \\ a & \text{if }\; i\neq j \end{cases}, $$ where the pairwise covariance $0<a<1$ is fixed and does not depend on $n$. Fix $t\in\mathbb{R}$ and let $$Y_n := \frac{\#\{i:X_i\geq t\}}{n}$$ be the proportion of $X_i$'s exceeding the threshold $t$. So $Y_n$ are discrete RVs taking values in $[0,1]$.
Question: What is the limiting distribution of $Y_n$ as $n\to\infty$?
Numerically $Y_n$ seem to converge (weakly) to a continuous distribution on $[0,1]$, and this behaviour is clearly different from the case of independent RVs governed by the CLT.
Below is the plot of the density of $Y_n$ for $a=0.1$, $n=10^4 \text{ and } 10^6$. The threshold $t$ is so that $\mathbb{E}Y_n=0.2$.
Solution by Roman Vershynin.
We can realize $X_i$ as $$ X_i = \sqrt{1-a}g_i + \sqrt{a}h $$ where $g_i$, $h$ are (jointly) independent $\mathcal{N}(0,1)$ random variables.
Condition on $h$. This makes $X_i$ (conditionally) independent. Thus, by the strong law of large numbers, for any value of $h$, we have $$ Y_n \xrightarrow{\text{a.s.}} \mathbb{P}\{\sqrt{1-a}g + \sqrt{a}h \geq t\mid h\} = \mathbb{P}\Bigl\{g \geq \frac{t - \sqrt{a}h}{\sqrt{1-a}} \mid h\Bigr\} = \Phi\Bigl(\frac{\sqrt{a}h - t}{\sqrt{1-a}}\Bigr) =: f(t, h) $$ where $\Phi$ is the standard normal cumulative distribution function.
Thus, for every fixed values $h$, $q$: $$ \mathbb{P}\{Y_n \leq q\mid h\} \to \mathbb{1}_{\{f(t, h) \leq q\}} $$
Notice that $f(t, h) \leq q$ iff $h\leq \frac{\sqrt{1-a}\Phi^{-1}(q)+t}{\sqrt{a}}$. Taking expectation w.r.t. $h$ on both sides, we get $$ \mathbb{P}\{Y_n \leq q\} \to \mathbb{E}_h \mathbb{1}_{\{f(t, h) \leq q\}} = \Phi\bigg(\frac{\sqrt{1-a}\Phi^{-1}(q)+t}{\sqrt{a}}\bigg) $$