= Solution
Take a <random variable> $Z$ with the <Rademacher distribution>, and set
$$
X_n=Z,\qquad Y_n=\begin{cases}Z,&n\text{ even},\\-Z,&n\text{ odd}.\end{cases}
$$
Both <marginal distributions> equal the law of $Z$ for every $n$. Hence $X_n$ and $Y_n$ each have <weak convergence of random variables> to $Z$; one may take $X=Y=Z$ on this <probability space>. But their sums alternate:
$$
X_n+Y_n=\begin{cases}2Z,&n\text{ even},\\0,&n\text{ odd}.\end{cases}
$$
Indeed, at $t=\pi/2$ their <characteristic functions> are $-1$ for even $n$ and $1$ for odd $n$, so the sums have no weak limit at all. \b[Separate marginal weak convergence does not imply convergence of the sum.] The <marginal weak convergence does not control sums> counterexample changes the dependence while leaving the marginals fixed. Unlike <convergence in probability>, separate marginal <weak convergence of random variables> contains no information about the joint coupling.
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