= Solution
The <permutation character> $\xi^\mu(g)$ is the number of $\mu$-<tabloids> fixed by $g$, since the trace of a permutation matrix counts fixed basis elements. It is therefore an integer. Arrange the partitions in decreasing lexicographic order. If $\lambda$ strictly dominates $\mu$, the first part at which they differ is larger for $\lambda$, so this ordering places $\lambda$ before $\mu$.
For the first partition $(n)$, there is just one <tabloid>, and $\chi^{(n)}=\xi^{(n)}=1_{S_n}$ is integer-valued. At a general partition $\mu$, the decomposition in (iii) can be rearranged as
$$
\chi^\mu=\xi^\mu-\sum_{\substack{\lambda\unrhd\mu\\\lambda\ne\mu}}a_{\lambda\mu}\chi^\lambda,\qquad a_{\lambda\mu}\in\mathbb Z_{\geq0}.
$$
By induction, all characters in the sum are already integer-valued; the <permutation character> and all coefficients are also integral. Thus $\chi^\mu$ is integer-valued. Finite induction proves that <symmetric-group characters are integer-valued>:
$$
\boxed{\chi^\lambda(g)\in\mathbb Z\qquad(g\in S_n,\ \lambda\vdash n).}
$$
The diagonal coefficient one is essential: it lets us solve for $\chi^\mu$ without division, preserving integrality.
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