From the first 539 positive integers, we select some such that their sum is at least one-third of the sum of the original numbers. How many numbers do we need to select for this?
The sum of the first 539 positive integers, according to the known formula for the sum of an arithmetic series, is:

$$
S_{539}=\frac{539 \cdot 540}{2}=145530
$$

To ensure that the sum of the selected numbers is at least one-third of the above sum and to select as few numbers as possible, we need to choose larger numbers.

Let the smallest of the selected numbers be $x$, and the others be $x+1, x+2, \ldots, 539$.

We want the sum of these, $S^{\prime}$, to be at least one-third of $S$, that is,

$$
48510=\frac{145530}{3} \leq S^{\prime}=\frac{n}{2}(x+539)
$$

where $n$ is the number of selected numbers. We know that in an arithmetic series $a_{n}=a_{1}+(n-1) d$, in our case $d=1$ and

$$
539=x+(n-1) \cdot 1
$$

from which $n=540-x$. Substituting this into (1) and rearranging:

$$
(540-x)(x+539) \geq 97020
$$

This leads us to the following quadratic inequality:

$$
x^{2}-x-194040 \leq 0
$$

The positive root of the corresponding equation is $x=441$; this is the largest solution to the inequality. If this is the smallest selected number, then 99 numbers need to be selected.