Sure, here is the translated text:

```
(40 points) Let integers $a_{1}, a_{2}, \cdots, a_{2019}$ satisfy
$$
\begin{aligned}
1= & a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{2019}=99 . \\
\text { Let } f= & \left(a_{1}^{2}+a_{2}^{2}+\cdots+a_{2019}^{2}\right)- \\
& \left(a_{1} a_{3}+a_{2} a_{4}+\cdots+a_{2017} a_{2019}\right) .
\end{aligned}
$$

Find the minimum value $f_{0}$ of $f$, and determine the number of arrays $\left(a_{1}, a_{2}, \cdots, a_{2019}\right)$ for which $f=f_{0}$.
```
Given the conditions,
$$
\begin{aligned}
2 f= & a_{1}^{2}+a_{2}^{2}+a_{2018}^{2}+a_{2019}^{2}+ \\
& \sum_{i=1}^{2017}\left(a_{i+2}-a_{i}\right)^{2} .
\end{aligned}
$$

Since \(a_{1}, a_{2}\) and \(a_{i+2}-a_{i} (i=1,2, \cdots, 2016)\) are all non-negative integers, we have
$$
a_{1}^{2} \geqslant a_{1}, \quad a_{2}^{2} \geqslant a_{2},
$$

and \(\left(a_{i+2}-a_{i}\right)^{2} \geqslant a_{i+2}-a_{i} (i=1,2, \cdots, 2016)\).
Thus,
$$
a_{1}^{2}+a_{2}^{2}+\sum_{i=1}^{2016}\left(a_{i+2}-a_{i}\right)^{2}
$$
$$
\begin{array}{l}
\geqslant a_{1}+a_{2}+\sum_{i=1}^{2016}\left(a_{i+2}-a_{i}\right) \\
=a_{2017}+a_{2018} .
\end{array}
$$

From equations (1) and (2), we get

Combining \(a_{2019}=99\) and \(a_{2018} \geqslant a_{2017}>0\), we have
$$
\begin{aligned}
f \geqslant & \frac{1}{2}\left(2 a_{2017}+\left(99-a_{2017}\right)^{2}+a_{2017}^{2}+99^{2}\right) \\
& =\left(a_{2017}-49\right)^{2}+7400 \geqslant 7400 . \\
& \text { Let } a_{1}=a_{2}=\cdots=a_{1920}=1, \\
& a_{1920+2 k-1}=a_{1920+2 k}=k(k=1,2, \cdots, 49), \\
& a_{2019}=99,
\end{aligned}
$$

Verification shows that all inequalities are satisfied with equality.
Thus, the minimum value of \(f\) is \(f_{0}=7400\).
Next, consider the equality conditions of equation (3).
In this case, \(a_{2017}=a_{2018}=49\), and all inequalities in equation (2) are satisfied with equality, i.e.,
$$
\begin{array}{l}
a_{1}=a_{2}=1, \\
a_{i+2}-a_{i} \in\{0,1\}(i=1,2, \cdots, 2016) .
\end{array}
$$

Thus, \(1=a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{2018}=49\), and for each \(k(1 \leqslant k \leqslant 49)\), there are at least two terms in \(a_{1}, a_{2}, \cdots, a_{2018}\) that equal \(k\). It is easy to verify that this is also a sufficient condition for equation (3) to hold with equality.

For each \(k(1 \leqslant k \leqslant 49)\), let the number of terms in \(a_{1}, a_{2}, \cdots, a_{2018}\) that equal \(k\) be \(1+n_{k}\), where \(n_{k}\) is a positive integer, and
$$
\begin{array}{l}
\left(1+n_{1}\right)+\left(1+n_{2}\right)+\cdots+\left(1+n_{49}\right)=2018 \\
\Rightarrow n_{1}+n_{2}+\cdots+n_{49}=1969 .
\end{array}
$$

The number of positive integer solutions \((n_{1}, n_{2}, \cdots, n_{49})\) to this equation is \(\mathrm{C}_{1968}^{48}\), and each solution uniquely corresponds to an array \((a_{1}, a_{2}, \cdots, a_{2019})\) that satisfies equation (4) with equality. Therefore, the number of arrays \((a_{1}, a_{2}, \cdots, a_{2019})\) such that \(f=f_{0}\) is \(\mathrm{C}_{1968}^{48}\).