Three positive integers sum to 2010, and the sum of their reciprocals is $\frac{1}{58}$. What are these numbers?
Let the three positive integers be $a, b, c$, such that $a+b+c=2010$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{58}$. Multiplying the second equation by the product of the denominators and rearranging gives $58(a b+b c+c a)-a b c=0$. Define the polynomial

$$
P(x)=(x-a)(x-b)(x-c)=x^{3}-(a+b+c) x^{2}+(a b+b c+c a) x-a b c
$$

The value of this polynomial at $x=58$, using the conditions of the problem, is

$$
P(58)=58^{3}-58^{2}(a+b+c)+58(a b+b c+c a)-a b c=58^{3}-2010 \cdot 58^{2}=-6566528
$$

On the other hand,

$$
P(58)=(58-a)(58-b)(58-c)
$$

Let $p=a-58, q=b-58, r=c-58$. From the above, we have

$$
p q r=6566528=2^{7} \cdot 29^{2} \cdot 61 \quad \text { and } \quad p+q+r=a+b+c-3 \cdot 58=1836
$$

We also know that $p, q, r$ are positive integers, since the reciprocals of $a, b, c$ are all less than $\frac{1}{58}$. In the following, we will consider two cases based on how many of the numbers $p, q, r$ are divisible by 29.

1. First, let $p$ be divisible by $29^{2}$, and neither $q$ nor $r$ be divisible by 29. Based on the size of the sum, we only need to consider two possibilities: $p=29^{2}=841$ or $p=2 \cdot 29^{2}=1682$. If $p=841$, then $q+r=995$, and one of $q$ and $r$ must be odd, i.e., 1 or 61. In either case, the other number is not a power of 2. If $p=1682$, then $q+r=154, q \cdot r=2^{6} \cdot 61$. The symmetric solution is $q=122, r=32$. Therefore, the original numbers in this case are 1740, 180, and 90.
2. We still need to examine the case where $p$ and $q$ are divisible by 29, and $r$ is not. If $p$ is divisible by 61, then $p=1769, q+r=67, q \cdot r=2^{7} \cdot 29$, and there is no solution. The same situation occurs if $q$ is divisible by 61. If 61 is a divisor of $r$, then $r$ can only be of the form $r=2^{k} \cdot 61$, where $0 \leq k \leq 4$. (Since $2^{5} \cdot 61$ is already greater than 1836.) From $p+q+r=1836$, we know that dividing by 29 leaves a remainder of 9. None of the $2^{k} \cdot 61$ satisfy this (3, 6, 12, 24, 19 are the remainders), so there is no solution in this case.

The three numbers we are looking for are $1740, 180, 90$, for which indeed

$$
\frac{1}{1740}+\frac{1}{180}+\frac{1}{90}=\frac{1}{58}
$$