24. If a positive integer $m$ has exactly four positive divisors (including 1 and itself), and the sum of these four positive divisors is $n$, then $n$ is called a "good number". In the set $\{2010, 2011, \cdots, 2019\}$, there are ( ) good numbers.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
24. A.

Let $m=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdots p_{r}^{\alpha_{r}}$, where $p_{i}$ is a prime number, $\alpha_{i} \in \mathbf{Z}_{+}(i=1,2, \cdots, r), p_{1}<p_{2}<\cdots<p_{r}$.
By the divisor count theorem, we know that $m$ has
$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{r}+1\right)
$$

divisors.
According to the problem, we have
$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdots\left(\alpha_{r}+1\right)=4 \text {. }
$$

Thus, $m=a b$ or $c^{3}(a, b, c$ are primes $)$.
When $m=c^{3}$, the sum of the four divisors of $m$ is $1+c+c^{2}+c^{3}$, substituting into the equation
$$
1+c+c^{2}+c^{3}=x(x \in\{2010,2011, \cdots, 2019\}),
$$

there is no prime solution.
When $m=a b$, the sum of the four divisors of $m$ is
$$
1+a+b+a b=(a+1)(b+1) \text {. }
$$

If one of $a, b$ is the even prime 2, without loss of generality, let $a=2$, then $b$ is an odd prime. Hence
$$
6 \mid(a+1)(b+1) \text {. }
$$

In the set $\{2010,2011, \cdots, 2019\}$, $2010, 2016$ are multiples of 6,
$$
\begin{array}{l}
b=\frac{2010}{3}-1=669, \\
b=\frac{2016}{3}-1=671
\end{array}
$$

neither of which is a prime, leading to a contradiction.
If $a, b$ are both odd primes, then
$$
2|(a+1), 2|(b+1), 4 \mid(a+1)(b+1) \text {. }
$$

In the set $\{2010,2011, \cdots, 2019\}$, the multiples of 4 are $2012, 2016$, where $2012=4 \times 503$, but $a, b$ are both odd primes, so $a+1, b+1$ are even numbers no less than 4, hence 2012 does not fit the problem; $2016=4 \times 504$, we have $a=3, b=503$ both being primes.

In summary, in the set $\{2010,2011, \cdots, 2019\}$, there is only one good number, which is 2016.