In the case where the $X$'s take finitely many values, you can prove what I claimed using the max-flow min cut theorem. I have not written it down, but this should be extendable to the general case.

Here's what I mean. Suppose you have two finite sets $A$ and $B$, equipped with probability measures $p$ and $q$. ($A$ and $B$ are traditionally called "boys" and "girls"). Let $K\subset A\times B$ (i.e. $(a,b)\in K$ if $a$ and $b$ know each other).

A coupling respecting $K$ is a measure $m$ on $A\times B$ such that $m(\{a\}\times B)=p(a)$, $m(A\times\{b\})=q(b)$ and $m(K)=1$. If $S\subset A$, define $\beta(S)=\{b\in B\colon\exists a\in A\text{ such that }(a,b)\in K\}$ and for $T\subset B$, $\alpha(T)=\{a\in A\colon\exists b\in B\text{ such that }(a,b)\in K\}$. By the max flow-min cut theorem, a coupling exists if and only if $q(\beta(S))\ge p(S)$ for each $S\subset A$ and $p(\alpha(T))\ge q(T)$ for each $T\subset B$.

To apply this to your problem, suppose that $X$ takes values $1,\ldots,n$ with probabilities $p_1,\ldots,p_n$. Let $A=\{(i,j)\colon 1\le i,j\le n\}$ and $B=\{1,\ldots,n\}$. Equip $A$ with the probability measure $p(i,j)=p_ip_j$. Let $K=\{((i,j),k)\colon i=k\text{ or }j=k\}$ (so that $\alpha(S)=\pi_1(S)\cup \pi_2(S)$, the union of the projections onto the coordinates and $\beta(S)=S\times\{1,\ldots,n\}\cup \{1,\ldots,n\}\times S$). Then applying the above criterion, you get that a coupling exists if and only if $(\sum_{i\in S}p_i)^2\le \sum_{i\in S}q_i\le 1-(1-\sum_{i\in S}p_i)^2)$ for each $S\subset \{1,\ldots,n\}$.

In the case where the $X$'s take finitely many values, you can prove what I claimed using the max-flow min cut theorem. I have not written it down, but this should be extendable to the general case.

Here's what I mean. Suppose you have two finite sets $A$ and $B$, equipped with probability measures $p$ and $q$. ($A$ and $B$ are traditionally called "boys" and "girls"). Let $K\subset A\times B$ (i.e. $(a,b)\in K$ if $a$ and $b$ know each other).

A coupling respecting $K$ is a measure $m$ on $A\times B$ such that $m(\{a\}\times B)=p(a)$, $m(A\times\{b\})=q(b)$ and $m(K)=1$. If $S\subset A$, define $\beta(S)=\{b\in B\colon\exists a\in S\text{ such that }(a,b)\in K\}$ and for $T\subset B$, $\alpha(T)=\{a\in A\colon\exists b\in T\text{ such that }(a,b)\in K\}$. By the max flow-min cut theorem, a coupling exists if and only if $q(\beta(S))\ge p(S)$ for each $S\subset A$ and $p(\alpha(T))\ge q(T)$ for each $T\subset B$.

To apply this to your problem, suppose that $X$ takes values $1,\ldots,n$ with probabilities $p_1,\ldots,p_n$. Let $A=\{(i,j)\colon 1\le i,j\le n\}$ and $B=\{1,\ldots,n\}$. Equip $A$ with the probability measure $p(i,j)=p_ip_j$. Let $K=\{((i,j),k)\colon i=k\text{ or }j=k\}$ (so that $\alpha(S)=\pi_1(S)\cup \pi_2(S)$, the union of the projections onto the coordinates and $\beta(S)=S\times\{1,\ldots,n\}\cup \{1,\ldots,n\}\times S$). Then applying the above criterion, you get that a coupling exists if and only if $(\sum_{i\in S}p_i)^2\le \sum_{i\in S}q_i\le 1-(1-\sum_{i\in S}p_i)^2)$ for each $S\subset \{1,\ldots,n\}$.