= Solution
Let $E=\{\alpha<\kappa:\operatorname{cf}(\alpha)=\omega\}$. This <set> is stationary: in any <club set>, choose a strictly increasing countable sequence and take its supremum, which lies in the <club set> and has <cofinality> $\omega$. For each $\alpha\in E$ choose an increasing cofinal sequence $c_\alpha:\omega\to\alpha$.
Fix $\beta<\kappa$. For every $\alpha\in E$ above $\beta$, some $c_\alpha(n)$ exceeds $\beta$. Partition this stationary tail by the least such $n$. A countable union of nonstationary <sets> is nonstationary, because fewer than $\kappa$ <club sets> have <club filter completeness>, so some cell is stationary. On it the <regressive function> $\alpha\mapsto c_\alpha(n)$ has, by <Fodor lemma>, a stationary fiber at a value $\gamma>\beta$.
Let $I_n=\{\gamma<\kappa:\{\alpha\in E:c_\alpha(n)=\gamma\}\text{ is stationary}\}$. The preceding argument says $\bigcup_n I_n$ is unbounded in $\kappa$. Since $\operatorname{cf}(\kappa)>\omega$, at least one $I_n$ is unbounded and therefore has size $\kappa$. Its fibers are pairwise disjoint <stationary sets>. Enumerate $\kappa$ of them as $T_\xi$, $\xi<\kappa$, and define $S_\xi=T_\xi$ for $\xi>0$, while
$$
S_0=T_0\cup\left(\kappa\setminus\bigcup_{\xi<\kappa}T_\xi\right).
$$
Adding a remainder preserves stationarity and introduces no overlap. Consequently
$$
\boxed{\kappa=\bigsqcup_{\xi<\kappa}S_\xi,\qquad S_\xi\text{ stationary for every }\xi.}
$$
This proves the <stationary partition by cofinal-sequence fibers> directly for every regular uncountable $\kappa$.
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