*(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. 

>**I. The Baby monster**

Just like the j-function $j(\tau)$ above and the Monster, the [modular function][1] $j_{2A}(\tau)$ that can be related to the *Baby monster*,

$$\begin{aligned}j_{2A}(\tau) &=\Big(\big(\tfrac{\eta(\tau)}{\eta(2\tau)}\big)^{12}+2^6 \big(\tfrac{\eta(2\tau)}{\eta(\tau)}\big)^{12}\Big)^2 \\
&=\sum\limits_{n=-1}^{\infty}a(n)q^n\\&= q^{-1} + \color{blue}{104} + 4372q + 96256q^2 + 1240002q^3+\cdots
\end{aligned}$$

also has,

$$\sum \limits_{n=1}^{24}a(n)^2\equiv 42 \;\; (\mathrm{mod} \;70),$$
 
Recall that $e^{\pi\sqrt{58}} =396^4-\color{blue}{104}.00000017\dots$.

>**II. Modular lambda function**

Given the [modular lambda function][2] $\lambda(\tau)=\lambda$ such that,

$$j(\tau) =  \frac{256(1-\lambda+\lambda^2)^3}{\lambda^2 (1-\lambda)^2}$$

and,

$$\begin{aligned}\lambda(\tau) &= \Big(\tfrac{\sqrt{2}\,\eta\big(\tfrac{\tau}{2}\big)\eta^2(2\tau)}{\eta^3(\tau)}\Big)^8\\
&=\sum\limits_{n=1}^{\infty}b(n)q^n\\
&=16q - 128q^2 + 704 q^3 - 3072q^4 + 11488q^5 - 38400q^6 + \dots
\end{aligned}$$

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 functions, there is a reason for the congruences other than whimsy.

>$\color{red}{Update}$

**P.S.** By a *really* serendipitous error almost straight from the pages of the *Hitchhiker's Guide*, it turns out the same $24$ coefficients $s(n)$ of *all* four sequences also obey,

$$\sum_{n=1}^{24} \frac{s(n)^2+42}{70} = \,\text{Integer}$$

The discussion below should illustrate how the error was found. 


 
  [1]: https://mathoverflow.net/questions/186336/
  [2]: https://en.wikipedia.org/wiki/Modular_lambda_function