(Too long for a comment.)
Since the authors point out their two observations in a "jocund" air, we can match their observations with another pair equally jocundly.
I. The Baby monster
Just like the j-function above and the Monster, the modular function that can be related to the Baby monster,
$$\begin{aligned}j_{2A}(\tau) &=\Big(\big(\tfrac{\eta(\tau)}{\eta(2\tau)}\big)^{12}+2^6 \big(\tfrac{\eta(2\tau)}{\eta(\tau)}\big)^{12}\Big)^2 \\ &=a(n)q^n\\&= q^{-1} + \color{blue}{104} + 4372q + 96256q^2 + 1240002q^3+\cdots \end{aligned}$$
also has,
$$\sum \limits_{n=1}^{24}a(n)^2\equiv -42 \;\; (\mathrm{mod} \;70),$$
Recall that $e^{\pi\sqrt{58}} =396^4-\color{blue}{104}.00000017\dots$.
(I checked the other monster-related functions in the link above and none other had the congruence.)
II. Modular lambda function
Given the modular lambda function,
$$\begin{aligned}\lambda(\tau) &= \left(\tfrac{\sqrt{2}\,\eta(\tfrac{\tau}{2})\eta^2(2\tau)}{\eta^3(\tau)}\right)^8\\ &=b(n)q^n\\ &=16q - 128q^2 + 704 q^3 - 3072q^4 + 11488q^5 - 38400q^6 + \dots \end{aligned}$$
then,
$$\sum \limits_{n=1}^{24}b(n)^2\equiv -42 \;\; (\mathrm{mod} \;70),$$
Of course, it is well-known that,
$$1^2+2^2+3^2+\dots+24^2 = 70^2$$
It should be interesting if, for these four related function, there is a reason for the congruences other than whimsy.