The only positive integers $n$ satisfying the equation $(1) \;\; \sum_{i=1}^3 \frac{n}{d_i} = 1457$. where $1 = d_1 < d_2 < \cdots < d_k=n$ are the divisors $n$, are $n=987,1023,1085,1175$. Observe that the sum on the LHS of equation (1) is the sum of the three largest divisors of $n$. Assume $n$ satisfies equation (1). Let $D(n)$ be the number of prime divisors of $n$. Let us consider the following two cases: Case 1: $D(n)=1$. Then there is a prime $p$ and an integer $k \geq 2$ s.t. $n=p^k$. Then according to equation (1) we obtain $(2) \;\; p^k + p^{k-1} + p^{k-2} = 1457$, If $k=2$, then $(2p + 1)^2 = 4 \cdot 1457 - 3 = 5828 - 3 = 5825 = 5^2 \cdot 233$, which implies $p \not \in \mathbb{N}$. Consequently $k>2$. Therefore $p \mid 1457 = 31 \cdot 47$ by equation (2), yielding $p \in \{31,47\}$, which according to equation (2) implies $1457 > p^k \geq p^3 \geq 31^3$, a contradiction which implies there is no solution of equation (1). Case 2: $D(n)>1$. Let $p$ and $q$ with $p<q$ be the two smallest prime divisors of $n$. Then $d_2=p$ and $d_3 \in \{p^2,q\}$. Assume $d_3=p^2$. Then $q > p^2$ and by equation (1) ${\textstyle (3) \;\; \frac{n}{p^2}(p^2 + p + 1) = 1457}$. Combining equation (3) with the facts that ${\textstyle \frac{n}{p^2} \in \mathbb{N}}$ and $31 \cdot 47$ is the prime factorization of $1457$ give us $p^2 + p + 1 \in \{31,47,1457\}$, yielding $(2p + 1)^2 \in \{11^2,5 \cdot 37, 5^2 \cdot 233\}$, implying $p=5$ and by equation (2) $n = \frac{31 \cdot 47 \cdot p^2}{p^2 + p + 1} = \frac{31 \cdot 47 \cdot 5^2}{5^2 + 5 + 1} = \frac{31 \cdot 47 \cdot 25}{31} = 47 \cdot 25 = 1175$, yielding $q = 47 > 5^2 = p^2$. Hence $n=1175$ is indeed a solution of equation (1). Finally assume $d_3=q$, which inserted in equation (1) result in ${\textstyle n(1 + \frac{1}{p} + \frac{1}{q}) = 1457}$, or alternatively ${\textstyle (4) \;\; \frac{n}{pq}(pq + p+ q) = 31 \cdot 47}$. The fact that ${\textstyle \frac{n}{pq} \in \mathbb{N}}$ (since $p$ and $q$ are distinct prime divisors of $n$) combined with equation (4) give us $pq + p + q \in \{31,47,31 \cdot 47\}$, i.e. $(4) \;\; (p + 1)(q + 1) = r \in \{32,48,1458\} = \{2^5,2^4 \cdot 3,2 \cdot 3^6\}$. If $r=2^5$, then $(p,q)=(3,7)$ and $n=47pq$. Likewise, if $r=2^4 \cdot 3$, then $(p,q)=(2,15),(3,11),(5,7)$ and $n=31pq$. If $r=2 \cdot 3^6$, then $(p + 1) + (q + 1)$ is odd, implying $p=2$ and $q=2 \cdot 3^5 - 1 = 2 \cdot 243 - 1 = 486 - 1 = 485$, which is impossible since $q$ is a prime. Summa summarum, since $p$ and $q$ are primes, we obtain following three solutions of equation (3): $n = 47 \cdot 3 \cdot 7 = 987$, $n = 31pq = 31 \cdot 3 \cdot 11 = 1023$ and $n = 31 \cdot 5 \cdot 7 = 1085$. Conclusion: Equation (1) has exactly four solutions, namely $n=987,1023,1085,1175$. q.e.d.