14. A unit spent 500,000 yuan to purchase a piece of high-tech equipment. According to the tracking survey of this model of equipment: after the equipment is put into use, if the maintenance and repair costs are averaged to the first day, the conclusion is: the maintenance and repair cost on the $x$-th day is
$$
\left[\frac{1}{4}(x-i)+5 \sim 0\right] \text { yuan. }
$$
(1) If the total maintenance and repair costs from the start of use to the scrapping of the equipment, as well as the cost of purchasing the equipment, are averaged to each day, this is called the daily average loss. Please express the daily average loss $y$ (yuan) as a function of the usage days $x$ (days);
(2) According to the technical and safety management requirements of this industry, when the average loss of this equipment reaches its minimum value, it should be scrapped. How many days should the equipment be used before it should be scrapped?

[Note]: In solving this problem, you may need to use the following two mathematical points (if needed, you can directly cite the following conclusions):
(A) For any integer $n$, the following equation must hold:
$$
1+2+3+4+\cdots+n=\frac{n(n+1)}{2} \text {. }
$$
(B) For any positive constants $a$ and $b$, and for any positive real number $x$, the following inequality must hold:
$\frac{a}{x}+\frac{x}{b} \geqslant 2 \sqrt{\frac{a x}{x b}}=2 \sqrt{\frac{a}{b}}$. It can be seen that $2 \sqrt{\frac{a}{b}}$ is a constant. That is, the function $y=\frac{a}{x}+\frac{x}{b}$ has a minimum value of $2 \sqrt{\frac{a}{b}}$, and this minimum value is achieved when $\frac{a}{x}=\frac{x}{b}$.
14. (1) The equipment is put into use for $x$ days. The average daily loss is:
$$
\begin{aligned}
y= & \frac{1}{x}\left[500000+\left(\frac{1}{4} \times 0+500\right)+\left(\frac{1}{4} \times 1+500\right)\right. \\
& \left.+\left(\frac{1}{4} \times 2+500\right)+\cdots+\left(\frac{x-1}{4}+500\right)\right] \\
= & \frac{1}{x}\left[500000+500 x+\frac{1}{4} \cdot \frac{x(x-1)}{2}\right] \\
= & \frac{500000}{x}+\frac{x}{8}+499 \frac{7}{8} .
\end{aligned}
$$
(2)
$$
\begin{aligned}
y & =\frac{5000000}{x}+\frac{x}{8}+499 \frac{7}{8} \\
& \geqslant 2 \sqrt{\frac{500000}{x} \times \frac{x}{8}}+499 \frac{7}{8} \\
& =500+499 \frac{7}{8}=999 \frac{7}{8} .
\end{aligned}
$$

When and only when $\frac{500000}{r}=\frac{x}{8}$, i.e., $x=2000$, the equality holds.
Therefore, the equipment should be scrapped after being put into use for 2000 days.