In Chapter 19 of Spivak's Calculus, Problem 36c)'s author-provided solution makes the following claim:
Supposing that $f$ is integrable on $[a,b]$, then, for any partition $\{t_0=a,t_1,\cdots,t_{n-1},t_n=b\}$ of $[a,b]$, there exists a smallest and largest sum of the form $\displaystyle \sum_{i=1}^kf(x_i)(t_i-t_{i-1})$ where $k \in [1,n]$.
I interpreted this as saying that, for a given $k$, I can find vectors $\vec{x}=\{x_1,x_2,\cdots,x_k\}$ and $\vec{z}=\{z_1,z_2,\cdots,z_k\}$ where $x_i,z_i \in [a,b]$ such that across all vectors $\vec{y}=\{y_1,y_2,\cdots,y_k\}$ where $y_i \in [a,b]$ we have:
$$\sum_{i=1}^k f(x_i)(t_i-t_{i-1}) \leq \sum_{i=1}^k f(y_i)(t_i-t_{i-1})\leq \sum_{i=1}^k f(z_i)(t_i-t_{i-1})$$
My book has not really committed that much time on series (those appear in the future chapters), so I am not certain how to proceed. It is clear that there is a lower bound and an upper bound to the collection of all possible sums $\sum_{i=1}^k f(y_i)(t_i-t_{i-1})$...however, I am not certain how to show that the lower and upper bound are actually in the image of the "sum function" $\sum_{i=1}^k f(y_i)(t_i-t_{i-1})$. In fact, this strikes me as intuitively false...but the book says otherwise.
Edit: I have assumed here that the members of $\vec x$ and $\vec z$ must all be distinct...but if this restriction is relaxed, then I suppose the claim is fine. I suspect, though, that we need to have distinct members.
It seems you misread your book(s). $x_1,\dots,x_n$ are fixed. The minimum and maximum values of $\sum_{i=1}^kf(x_i)(t_i-t_{i-1})$ are with respect to $k\in[1,n].$