= Solution
For permutation characters, <Burnside lemma> on a product gives
$$
\langle\pi_k,\pi_l\rangle_G
=\#\bigl(G\backslash(X_k\times X_l)\bigr).
$$
Two pairs $(A,B)$ and $(A',B')$ lie in the same $S_n$-orbit exactly when
$$
|A\cap B|=|A'\cap B'|.
$$
When $0\leq l\leq k\leq n/2$, every value $0,1,\ldots,l$ is possible because $k+l\leq n$. Therefore the <symmetric-group subset permutation representation> satisfies
$$
\boxed{\langle\pi_k,\pi_l\rangle_G=l+1.}
$$
For $1\leq r\leq n/2$, the <inclusion map between subset permutation modules> embeds $\mathbb C[X_{r-1}]$ into $\mathbb C[X_r]$, so
$$
\chi_r=\pi_r-\pi_{r-1}
$$
is a character. Its norm is
$$
\begin{aligned}
\langle\chi_r,\chi_r\rangle
&=\langle\pi_r,\pi_r\rangle
-2\langle\pi_r,\pi_{r-1}\rangle
+\langle\pi_{r-1},\pi_{r-1}\rangle\\
&=(r+1)-2r+r=1.
\end{aligned}
$$
By <character orthogonality>, $\chi_r$ is the character of an irreducible representation.
For $r>n/2$, complementation is an $S_n$-equivariant bijection $X_r\cong X_{n-r}$, so
$$
\pi_r=\pi_{n-r}.
$$
Consequently
$$
\pi_r-\pi_{r-1}
=-\bigl(\pi_{n-r+1}-\pi_{n-r}\bigr),
$$
the negative of one of the irreducible characters just found, and is not itself the character of a representation.
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