It seems to me that Etingof et al (pages 51-52) and Kopp and Wiltshire-Gordon give very similar proofs. I'll write them out in as elementary a way as I can, and also give a third proof of my own along similar lines.
Notation: Let $G$ be our finite group, $V$ an irrep over the complex numbers, $\rho_V$ the map $G \to GL(V)$ and $\chi_V$ the character of $V$. We write $\mathrm{Id}_V$ for the identity map $V \to V$. Let $g_1$, $g_2$, ..., $g_c$ be representatives for the conjugacy classes of $G$, and let $C(g)$ be the conjugacy class of $g$.
Lemma: For any $g \in G$, $\sum_{h \in C(g)} \rho_V(h) = \frac{|C(g)| \chi_V(g)}{\dim V} \mathrm{Id}_V \quad\quad (\ast)$
Proof: For any $f$ in $G$, we have $\rho_V(f) \cdot \left( \sum_{h \in C(g)} \rho_V(h) \right) = \sum_{h \in C(g)} \rho_V(fh) =$ $\left( \sum_{h \in C(g)} \rho_V(fhf^{-1}) \right) \cdot \rho_V(f) = \left( \sum_{h \in C(g)} \rho_V(h) \right) \cdot \rho_V(f).$
So the left hand side of $(\ast)$ commutes with every $\rho_V(f)$. By Schur's lemma, this means that the left hand side of $(\ast)$ is $a \mathrm{Id}_V$ for some scalar $a$. Taking traces, we compute that $|C(g)| \chi_V(g) = a \dim V$, so $a = |C(g)| \chi_V(g)/\dim V$ as required. $\square$
Let's define $P(g) = \sum_{h \in C(g)} h.$ This is an element in $\mathbb{Z}[G]$. Let $P_V(g) = \sum_{h \in C(g)} \rho_V(h).$ This is how $P(g)$ acts on $V$. So the above Lemma shows that $P_V(g) = |C(g)| \chi_V(g)/ \dim V \cdot \mathrm{Id}_V$.
Both the papers I cite want to sum up $P_V(g_i)$ in some way, use the identity $\sum_{i} |C(g_i)| \chi_V(g_i) \chi_V(g_i^{-1}) = |G| \quad \quad (\ast \ast)$ and wind up with $\dim V$ in the denominator of something which they can prove, by other means, is an integer. Note that you may recognize $(\ast \ast)$ better in the form $\sum_{g \in G} \chi_V(g) \overline{\chi_V(g)} = |G|$; we get identity $(\ast \ast)$ by grouping together the terms in the same conjugacy class and using $\chi_V(g^{-1}) = \overline{\chi_V(g)}$.
Etingof et al's proof: Consider $Q_V = \sum_i P_V(g_i) \chi_V(g_i^{-1}).$ On the one hand, using the Lemma, $Q_V = \sum_i |C(g_i)| \chi_V(g_i) \chi_V(g_i^{-1}) \frac{1}{\dim V} \mathrm{Id}_V = \frac{|G|}{\dim V} \mathrm{Id}_V$ using $(\ast\ast)$.
On the other hand if you expand out $Q_V$ in $\mathbb{C}[G]$, you'll see that the coefficient of every group element is an algebraic integer. So $|G|/\dim V$ is an algebraic integer and, since it is rational, must be an integer. $\square$.
Kopp and Wiltshire-Gordon's proof: Set $R = \sum_{i} \frac{|G|}{|C(g)|} P(g_i) P(g_i^{-1}).$ Let $R_V = \rho_V(R)$.
Using the Lemma, $R_V = \sum_i \frac{|G|}{(\dim V)^2} |C(g_i)| \chi_V(g_i) \chi_V(g_i^{-1}) \mathrm{Id} = \frac{|G|^2}{(\dim V)^2} \mathrm{Id}_V.$
So, similarly, $\rho_V(R^k) = \frac{|G|^{2k}}{(\dim V)^{2k}} \mathrm{Id}_V.$
Let $U$ be the regular representation of $G$. So $\chi_U = \sum (\dim V) \chi_V$. We deduce that $\chi_U(R^k) = \sum_V (\dim V) \frac{|G|^{2k}}{(\dim V)^{2k}} \mathrm{Tr}(\mathrm{Id}_V) = \sum_V (\dim V)^2 \frac{|G|^{2k}}{(\dim V)^{2k}}.$
One the other hand, $R$ is clearly in $\mathbb{Z}[G]$. (Recall that $|C(g)|$ divides $|G|$ because $|C(g)|$ has a transitive action of $G$ by conjugation.) So $R^k$ is in $\mathbb{Z}[G]$. And the trace of any element of $\mathbb{Z}[G]$ acting on the regular rep is an integer (in fact, it is $|G|$ times the coefficient of the identity). So $\sum_V (\dim V)^2 \frac{|G|^{2k}}{(\dim V)^{2k}}$ is an integer (in fact, one divisible by $|G|$) for all $k$. In Lemma 6.2, the authors show that this forces each $|G|/\dim V$ to be an integer. $\square$
Remark Other parts of the paper make use of the pleasant identity $R = \sum_{g \in g} \sum_{h \in G} ghg^{-1} h^{-1}$. But this doesn't seem to be important if our goal is solely to get this fact.
Variant on the second proof If one considers the action of $R$ on $\mathbb{Z} G$, it is clearly given by a matrix with integer entries. Let $f(\lambda)$ be the characterisitc polynomial of that matrix, so $f$ is a monic polynomial with rational coefficients.
We showed that $R$ acts on the subspace $V^{\oplus \dim V}$ of $\mathbb{C} G$ by $(|G|/\dim V)^2$. So $(|G|/\dim V)^2$ is an eigenvalue of the above integer matrix. So $(|G|/\dim V)^2$ is a root of $f$. By the rational root theorem, we deduce that $(\dim V)^2$ divides $|G|^2$, so $\dim V$ divides $|G|$. $\square$