How many positive integers less than 2018 can be expressed exactly as the sum of its three distinct positive divisors?
1. 336.

Let $n=a+b+c$, where $a>b>c>0$ and are divisors of $n$. Then,
$$
\frac{a}{n}+\frac{b}{n}+\frac{c}{n}=1.
$$

Simplifying the above equation to its simplest form, each numerator is 1 if and only if 1 can be expressed as the sum of three fractions in the following three ways, i.e.,
$$
\begin{array}{l}
1=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=\frac{1}{2}+\frac{1}{4}+\frac{1}{4} \\
=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}.
\end{array}
$$

Clearly, only the last expression meets the condition that the three denominators are all different.
Thus, $n=6k, a=3k, b=2k, c=k\left(k \in \mathbf{Z}_{+}\right)$.
Since $2018=6 \times 336+2$, $k$ can be any integer from 1, 2, ..., 335, 336.