$\quad$ $n=24k-1\equiv 2\pmod{3}\Longrightarrow n$ isn't a perfect square. Let $d,D\in\mathbb{N}$ such that $d\cdot D=n$. $\quad$ Results: a) $d\ne D$; b) we can split the positive divisors of $n$ into pairs $(d_i,D_i), i\in\{1,2,\cdots,m\}, m\in\mathbb{N}$ such that $d_i, D_i\in\mathbb{N}; d_i<D_i; d_i\cdot D_i=n$. Also, the sum of the positive divisors of $n$ is $N=\sum_{i=1}^m(d_i+D_i)$. Working in modulo 8: $d_iD_i=n\equiv 7\pmod{8}\Longrightarrow$ $\Longrightarrow d_i\equiv 1\pmod{8}, D_i\equiv 7\pmod{8}$ or $d_i\equiv 7\pmod{8}, D_i\equiv 1\pmod{8}$ or $d_i\equiv 3\pmod{8}, D_i\equiv 5\pmod{8}$ or $d_i\equiv 5\pmod{8}, D_i\equiv 3\pmod{8}$. In all situations, $d_i+D_i\equiv 0\pmod{8}\Longrightarrow 8|d_i+D_i$. Working in modulo 3: $d_iD_i=n\equiv 2\pmod{3}\Longrightarrow$ $\Longrightarrow d_i\equiv 1\pmod{3}, D_i\equiv 2\pmod{3}$ or $ d_i\equiv 2\pmod{3}, D_i\equiv 1\pmod{3}$. Results: $d_i+D_i\equiv 0\pmod{3}\Longrightarrow 3|d_i+D_i$. $8|d_i+D_i$ and $3|d_i+D_i\Longrightarrow 24|d_i+D_i\Longrightarrow$ $24|\sum_{i=1}^m(d_i+D_i)=N$.

$\displaystyle \sum _{d|n} d=\sum _{d|n,\ d^{2} < n}\left( d+\frac{n}{d}\right) =\sum _{d|n,\ d^{2} < n}\left(\frac{d^{2} +n}{d}\right)$ (Note that $n$ cannot be a perfect square) Since $\displaystyle 24|d^{2} +n $ and $\displaystyle \gcd( d,24) =1$, sum of all positive divisors of $\displaystyle n$ is divisible by 24

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