= Solution
Write $P=999/1000$. If
$$
t>P\quad\Longleftrightarrow\quad
\frac n{n+3}>\frac{999}{1000}
\quad\Longleftrightarrow\quad
\boxed{n>2997},
$$
then both players prefer buying against a buyer, since $(t-P)/2>0$. Both prefer not buying against a nonbuyer: $2t/3<2/3<P$ and $t/3<1/3<P$. These strict opposite <best responses> give the interior crossing of a <coordination game>.
An explicit <fully mixed Nash equilibrium> follows from indifference. Let $p$ be X's <probability> of buying and $q$ Y's. X is indifferent when
$$
(1-q)\left(\frac{2t}{3}-P\right)+q\frac{t-P}{2}=0,
$$
and Y is indifferent when
$$
(1-p)\left(\frac t3-P\right)+p\frac{t-P}{2}=0.
$$
Solving gives
$$
\boxed{
p=\frac{2(P-t/3)}{P+t/3},\qquad
q=\frac{2(P-2t/3)}{P-t/3},\qquad t=\frac n{n+3},\ n>2997.
}
$$
Both numerators and denominators are positive. For $p<1$ the required inequality is $2P-2t/3<P+t/3$, equivalent to $P<t$; the same comparison for $q<1$ again reduces to $P<t$. Thus $p,q\in(0,1)$, and the equations make both actions <best responses> for each player. By linearity of <expected payoff>, neither player can profit by any other <mixed strategy>. At $n=2997$ these formulas reach $p=q=1$ and are not fully mixed; the strict threshold matters.
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