Solution:
By virtue of $z \geq 1675$ we have
$$y+z<2015 \Leftrightarrow y<2015-z \leq 2015-1675<1675$$
It follows that $(1675-y) \cdot (1675-z) \leq 0 \Leftrightarrow y \cdot z \leq 1675 \cdot (y+z-1675)$.
By using the inequality $u \cdot v \leq \left(\frac{u+v}{2}\right)^2$ for all real numbers $u, v$ we obtain
$$\begin{gathered} P = x \cdot y \cdot z \leq 1675 \cdot x \cdot (y+z-1675) \leq 1675 \cdot \left(\frac{x+y+z-1675}{2}\right)^2 = \\ 1675 \cdot \left(\frac{2015-1675}{2}\right)^2 = 1675 \cdot 170^2 = 48407500 \end{gathered}$$
$$\text{We have } P = x \cdot y \cdot z = 48407500 \Leftrightarrow \left\{\begin{array}{l} x + y + z = 2015, \\ z = 1675, \\ x = y + z - 1675 \end{array}\right. \Leftrightarrow \left\{\begin{array}{l} x = 170 \\ y = 170 \\ z = 1675 \end{array}\right.$$
So, the greatest value of the product is $P = x \cdot y \cdot z = 48407500$.

Let $S = \{(x, y, z) \mid x \geq 20, y \geq 40, z \geq 1675, x+y+z=2015\}$ and $\Pi = \{|x \cdot y \cdot z| \mid (x, y, z) \in S\}$. We have to find the biggest element of $\Pi$. By using the given inequalities we obtain:
$$\left\{\begin{array}{l} 20 \leq x \leq 300 \\ 40 \leq y \leq 320 \\ 1675 \leq z \leq 1955 \\ y < 1000 < z \end{array}\right.$$
Let $z = 1675 + d$. Since $x \leq 300$ so $(1675 + d) \cdot x = 1675 x + d x \leq 1675 x + 1675 d = 1675 \cdot (x + d)$. That means that if $(x, y, 1675 + d) \in S$ then $(x + d, y, 1675) \in S$, and $x \cdot y \cdot (1675 + d) \leq (x + d) \cdot y \cdot 1675$. Therefore $z = 1675$ must be for the greatest product.
Furthermore, $x \cdot y \leq \left(\frac{x+y}{2}\right)^2 = \left(\frac{2015-1675}{2}\right)^2 = \left(\frac{340}{2}\right)^2 = 170^2$. Since $(170, 170, 1675) \in S$ that means that the biggest element of $\Pi$ is $170 \cdot 170 \cdot 1675 = 48407500$