= Solution
Put $p_i=\mathbb P(X_i=1)$ and $\lambda=\sum_i p_i$. A relative-entropy form of the <Poisson approximation bound for dependent Bernoulli variables> is
$$
D_e(P_{S_n}\Vert\operatorname{Poisson}(\lambda))
\leq\sum_{i=1}^np_i^2
+\sum_{i=1}^nH_e(X_i)-H_e(X_1,\ldots,X_n).
$$
The last two terms form the <total correlation>; they vanish when the Bernoulli variables are independent.
Here $D_e$ and $H_e$ use <natural logarithms>.
To prove the bound, let $Q_i$ be the <Poisson distribution> with mean $p_i$ and let $Q=\bigotimes_iQ_i$. Expanding the <Kullback-Leibler divergence> against this product law gives
$$
D_e(P_{X_1^n}\Vert Q)
=\sum_iD_e(\operatorname{Bernoulli}(p_i)\Vert\operatorname{Poisson}(p_i))
+\sum_iH_e(X_i)-H_e(X_1^n).
$$
The supplied one-dimensional estimate bounds the first sum by $\sum_i p_i^2$. Under the addition map, $P_{X_1^n}$ becomes $P_{S_n}$, while the <sum of independent Poisson random variables> under $Q$ has the Poisson distribution with mean $\lambda$. The <data processing inequality for relative entropy> proves the displayed result.
If a bound directly in the paper's unhalved total-variation norm is desired, <Pinsker's inequality> also gives
$$
\lVert P_{S_n}-\operatorname{Poisson}(\lambda)\rVert_1
\leq\sqrt{2\left\{\sum_i p_i^2+\sum_iH_e(X_i)-H_e(X_1^n)\right\}}.
$$
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