Hint 1: Argue, where the break points happen (i.e., let, $x_{n_1},\dots,x_{n_{25}}$ are the locations, where, $x_{n_i}\neq x_{n_i-1}$. Analyze, how it behaves between the break points). Hint 2: $25\cdot 26-25-2=623$.

See that if $x_1>1$, then we can replace $A=\{x_1,\cdots,x_{2017}\}$ with $A_0=\{x_1-1,\cdots,x_{2017}-1\}$. This will make the sum smaller. Hence, we can assume that $x_1=1$. Now, we can replace $2017$ with any sufficiently large number $n$ by simply adding $1$s between $x_0$ and $x_1$. See that if there exist a number $1\leq i\leq n-1$ such that $x_{i+1}>x_i+1$, then we can replace $A=\{x_1,\cdots,x_n\}$ with $A_0=\{x_1,\cdots, x_i, x_{i+1}-1, x_{i+2}-1,\cdots, x_n-1\}$. This will make the sum smaller. Hence, we can assume that the set $A$ consists every integer from $1$ to $25$. By similar observations, we can also assume that the set $A$ doesn't contain any integer greater than $25$. Claim: For every integer $k\in\{1,2,3,\cdots ,24\}$; one can assume that at least two of $x_i$s are equal to $k$. Proof: Let's prove this by induction over $k$. For $k=1$, we have $x_0=x_1=1$, done. Assume that $x_{j-1}=x_j=m, x_{j+1}=m+1$ where $m\in [1,23]$. Let's show that $x_{j+2}$ must be $m+1$. Suppose that $x_{j+2}=m+2$. Then, add a new number $x_{new}=m+1$ between $x_{j+1}$ and $x_{j+2}$. (i.e. replace $A=\{x_1,\cdots, x_{j+1}, x_{j+2},\cdots ,x_n\}$ with $A=\{x_1,\cdots, x_{j+1}, x_{new}, x_{j+2},\cdots ,x_n\}$) See that this process makes the sum smaller. This finishes the proof of the claim. Now let's calculate the sum. For every $i\in [2,25]$, let $a_i$ be the smallest index such that $x_{a_i}=i$. By our Claim, we know that $x_{a_i+1}=i$ for all $i\in [2,24]$. Also, by our Claim, $x_{a_i-1}=x_{a_i-2}=i-1$ for all $i\in [2,25]$. Then, $$(x_{a_i}-x_{a_i-2})x_{a_i}=i\text{ for all }i\in[2,25]$$$$(x_{a_i+1}-x_{a_i-1})x_{a_i+1}=i\text{ for all }i\in[2,24]$$Hence, $\sum_{i=2}^{n}(x_i-x_{i-2})x_i\ge (2+3+\cdots +24+25)+(2+3+\cdots +24)=623$, as desired. For equality, every number from $1$ to $24$ should appear at least twice and it is easy to see that there should be exactly one $x_i$ which is equal to $25$. Every set satisfying these conditions gives us equality. For all $i\in [1,24]$, let $t_i$ be the number of $x_j$s which are equal to $i$. ($0\leq j\leq 2017$) We have $t_1+\cdots +t_{24}=2017$ and $t_i\ge 2$. If we let $r_i=t_i-2$, then $r_1+\cdots +r_{24}=1969$ and $r_i\ge 0$. We know that the number of sequences satisfying this conditions is $\binom{1969+24-1}{24-1}=\binom{1992}{23}$. Hence, the number of sequences that hold the case of equality is $\dbinom{1992}{23}$.