= Solution
We use the standard facts that finite <complex representations> are completely reducible, <irreducible characters> are orthonormal, the regular character is $\sum_\chi\chi(1)\chi$, and the number of <irreducible characters> equals the number of conjugacy classes. We also use Schur's lemma, Sylow's theorem and the nontrivial center of a nontrivial finite $p$-group.
For an <irreducible character> $\chi$ of degree $d$, a <conjugacy-class sum> $z_C$ acts as the scalar $\lambda_C=|C|\chi(g)/d$ by Schur's lemma and taking <traces>. This scalar is an <algebraic integer>: class sums form an <integral> basis of the center of the <integral> <group algebra>, their multiplication has <integer> structure constants, and multiplication by $z_C$ satisfies its monic <integer> characteristic <polynomial>. That same <polynomial> annihilates its image in any <irreducible representation>. Character values are also <algebraic integers>, since they are sums of roots of unity. <Character orthogonality> now gives
$$
\frac{|G|}{d}=\sum_C\frac{|C|\chi(g)}d\overline{\chi(g)}=\sum_C\lambda_C\overline{\chi(g)}.
$$
This is a rational <algebraic integer>, hence an <integer>. Therefore \b[$\chi(1)$ divides $|G|$].
The <Burnside's pa qb theorem> states that every <finite group> of order $p^a q^b$ is soluble. We prove its character-theoretic ingredient. If a class has size $m$ coprime to $d$, <integers> $u,v$ with $ud+vm=1$ give
$$
\frac{\chi(g)}d=u\chi(g)+v\frac{m\chi(g)}d,
$$
an <algebraic integer>. Every <algebraic conjugate> of this number is the average of $d$ roots of unity and has modulus at most one. If it is nonzero, its algebraic norm is a nonzero <integer> of absolute value at least one; since every factor has modulus at most one, all must have modulus one. In particular $|\chi(g)|=d$. Equality in the <triangle inequality> for the unit-modulus <eigenvalues> implies that $\rho(g)$ is scalar. Thus \b[if $(m,d)=1$, either $\chi(g)=0$ or $\rho(g)$ is scalar].
Suppose a <group> of order $p^a q^b$ were nonabelian simple, with both <primes> present. Choose a nonidentity $g$ in the center of a Sylow $p$-subgroup. Its centralizer contains that <Sylow subgroup>, so its class size is a power of $q$. Every nontrivial <irreducible representation> is faithful by simplicity. If it sent $g$ to a scalar, faithfulness would make $g$ central in the <group>, impossible in a nonabelian simple <group>. The ingredient above therefore gives $\chi(g)=0$ for every nontrivial character with degree coprime to $q$. Evaluating the regular character at $g\ne1$ gives
$$
0=1+\sum_{q\mid\chi(1)}\chi(1)\chi(g),\qquad
\frac1q=-\sum_{q\mid\chi(1)}\frac{\chi(1)}q\chi(g).
$$
The right side is an <algebraic integer>, but $1/q$ is not, a contradiction. A <group> of prime-power order cannot be nonabelian simple either, by its nontrivial center. Induct on the order: every nonsimple <group> has a proper nontrivial <normal subgroup>, and that subgroup and the quotient have smaller orders with at most these two <prime> divisors. By induction both are soluble; their derived-series extensions make the original <group> soluble. Simple <abelian groups> have <prime> order and start the induction. This proves \b[Burnside's theorem].
For the final divisibility, let $|G|=n$ be odd and let $k$ be its number of conjugacy classes. All irreducible degrees are odd by the divisibility just proved. We use the standard <Frobenius-Schur indicator> theorem: $\nu(\chi)=n^{-1}\sum_g\chi(g^2)$ is zero precisely for an <irreducible character> not equal to its conjugate, and is $\pm1$ otherwise. Squaring is a bijection on an odd-order <group>: for each $g$, its unique square root is $g^{(n+1)/2}$, since every element order divides $n$. For every nontrivial <irreducible character>, orthogonality to the trivial character consequently gives $\nu(\chi)=n^{-1}\sum_g\chi(g)=0$. The nontrivial characters therefore occur in distinct conjugate pairs of equal odd degree. Since an odd square is one modulo eight,
$$
n-k=\sum_\chi(\chi(1)^2-1)
$$
is a sum of pairs $2(d^2-1)$, each divisible by sixteen; the trivial character contributes zero. Hence
$$
\boxed{16\mid n-k.}
$$
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