This is not an answer, but an extended comment, that should help anyone interested in using brute-force numerical search to find the solution.

The problem is to find $n \in \mathbb{N}$ for which
$$7921 \sigma(n) = 15840 n \tag{1}\label{NA1}$$
where $\sigma(n)$ is the sum of all divisors of $n$, as defined in the Wikipedia divisor function article, and as a sequence in OEIS A000203.

Because $\sigma(n) = n + 1$ if $n$ is a prime, and
$$7921 (n + 1) \ne 15840 n, \quad n \in \mathbb{N}$$
we already know there is no prime solution to $\eqref{NA1}$.

Consider the prime factorization of $n$. Let $p_i \in \mathbb{N}$ be nonrepeating primes ($p_i = p_j$ if and only if $i = j$), and $1 \le k_i \in \mathbb{N}$. Then,
$$n = \prod_{i=0}^{N-1} p_i^{k_i}$$
and
$$\sigma(n) = \prod_{i=0}^{N-1} \frac{p_i^{k_i+1} - 1}{p_i - 1}$$
because $\sigma(p^k) = \sum_{j=0}^{k} p^j = (p^{k+1}-1)/(p-1)$ when $p$ is a prime.

We can now rewrite the problem $\eqref{NA1}$ as
$$7921 \prod_{i=0}^{N-1} \frac{p_i^{k_i+1} - 1}{p_i - 1} = 15840 \prod_{i=0}^{N-1} p_i^{k_i} \tag{2}\label{NA2}$$
Rearranging the terms yields
$$\prod_{i=0}^{N-1} \frac{ p_i^{k_i + 1} - p_i^{k_i} }{ p_i^{k_i + 1} - 1 } = \frac{7921}{15840} = \frac{n}{\sigma(n)} = \frac{89^2}{2^5 \cdot 3^2 \cdot 5 \cdot 11}
\tag{3}\label{NA3}$$
Note the term
$$f_i = \frac{ p_i^{k_i + 1} - p_i^{k_i} }{ p_i^{k_i + 1} - 1 } = \frac{p_i^{k_i}}{\sum_{j=0}^{k_i} p^j}, \quad \frac{1}{2} \lt f_i \lt 1
\tag{4}\label{NA4}$$
i.e.,
$$\begin{array}{ll}
f_i = \frac{p_i}{p_i + 1}, & k_i = 1 \\
f_i = \frac{p_i^2}{p_i^2 + p_i + 1}, & k_i = 2 \\
f_i = \frac{p_i^3}{p_i^3 + p_i^2 + p_i + 1 }, & k_i = 3 \\
f_i = \frac{p_i^{k_i}}{p_i^{k_i} + p_i^{k_i-1} + \dots + p_i + 1 } & \\
\end{array}$$

Thus, the numerical search problem is now reduced to find the set of terms $f_i$ based on primes $p_i$ and their positive powers $k_i$, so that the product $$\prod_{i=0}^{N-1} f_i = \frac{7921}{15840}$$
In particular, because $f_i \lt 1$, a particular set can be rejected immediately if the product falls below the target ratio.

For example, if $p_0 = 89$, $k_0 = 2$, to eliminate the prime factor in the numerator. Repeating, that leads to $p_1 = 8011$, $k_1 = 1$; $p_2 = 2003$, $k_2 = 1$; and $p_3 = 167$, $k_3 = 1$, to get us to a result with a composite numerator and a denominator:
$$\begin{array}{r|l|l}
n & \frac{n}{\sigma(n)} & \frac{15840 n}{7921 \sigma(n)} \\
\hline
89^2 \cdot 8011 \cdot 2003 \cdot 167
& \frac{7921}{8064} = \frac{7921}{2^7 \cdot 3^2 \cdot 7}
& \frac{55}{28} = \frac{5 \cdot 11}{2^2 \cdot 7}
\\ 89^2 \cdot 8011 \cdot 2003 \cdot 167 \cdot 7
& \frac{7921}{9216} = \frac{7921}{2^{10} \cdot 3^2}
& \frac{55}{32} = \frac{5 \cdot 11}{2^5}
\\ 89^2 \cdot 8011 \cdot 2003 \cdot 167 \cdot 2
& \frac{7921}{12096} = \frac{7921}{2^6 \cdot 3^3 \cdot 7}
& \frac{55}{42} = \frac{5 \cdot 11}{2 \cdot 3 \cdot 7}
\\ 89^2 \cdot 8011 \cdot 2003 \cdot 167 \cdot 2 \cdot 7
& \frac{7921}{13824} = \frac{7921}{2^9 \cdot 3^3}
& \frac{55}{48} = \frac{5 \cdot 11}{2^4 \cdot 3}
\\ 89^2 \cdot 8011 \cdot 2003 \cdot 167 \cdot 2^2
& \frac{7921}{14112} = \frac{7921}{2^5 \cdot 3^2 \cdot 7^2}
& \frac{55}{49} = \frac{5 \cdot 11}{7^2}
\\ \end{array}$$
If you append $p_5 = 7$, $k_5 = 1$ or $k_5 = 2$ to $n$ in the final row above, the rightmost field drops below 1 (to $55/56$ for $k_5 = 1$, and to $55/57$ for $k_5 = 2$), leading nowhere. Similarly, appending $p_6 = 3$, $k_6 = 1$ or $k_6 = 2$ to $n$ in the second-to-last row (to $55/64$ for $k_6 = 1$, and to $165/208$ for $k_6 = 2$) leads nowhere.

It looks to (very non-mathematician) me that an exhaustive search over primes $p$ is possible, due to terms $f_i$ having a power of a prime in the numerator, as specified in $\eqref{NA4}$. Whether an exhaustive search is possible or not is an open question (and is important for those looking for proof), but the efficient numerical brute force search strategies are straightforward; especially if one is looking for some other ratios than $\frac{n}{\sigma(n)} = \frac{7921}{15840}$.

To continue the search above, I'd need a prime $p$ and a positive integer $k$ such that $\sum_{j=0}^{k} p^j = 55$ (to yield a factor with denominator $55$). No such pair exists, so the search strategies I've come up with thus far are exhausted.

Hopefully, one of the math sages here can take this further from here.