We will prove that the required maximum is attained when $x_i=0$ or $x_i=1\,,\, i=1,2,\dots,n$. The proof is based on the following two claims. Claim 1. Let us denote: \[ f_{j,r}(x) := \sum_{k=j}^{j+r-1}f\left ( \left | x-\frac{2k-1}{2n} \right | \right )\] and $I \subset [0,1]$ be an interval with $|I|\geq r/n$. Then $\max_{x\in I} f_{j,r}(x)$ is attained at some endpoint of $I$. The proof is a straightforward comparison using the fact that $f$ is increasing. Claim 2. Let $0\leq x_1 < x_2 <\dots < x_j $, where every knot $x_i$ has multiplicity $k_i$, with $k_1+k_2+\dots+k_j=n$. Then there exist $i\,,\, 1\leq i \leq j$, such that $x_{i+1}-x_{i-1} \geq k_i/n$, where for convenience we set $x_0=0, x_{j+1}=1$. Proof: Suppose on the contrary $x_{i+1}-x_{i-1} < k_i/n\,, i=1,2,\dots,j$. Summing up all these inequalities, we get a contradiction $\blacksquare$ Denote: \[\phi(x_1,x_2,\dots,x_n) = \sum_{k=1}^{n}f\left ( \left | x_k-\frac{2k-1}{2n} \right | \right )\] Suppose $\{x_1,x_2,\dots, x_n\}=\{y_1,y_2,\dots, y_j\}\,,\,y_1<y_2<\dots<y_j$, and each $y_i$ has multiplicity $k_i$. Using Claim 2, we take $y_i$ with $y_{i+1}-y_{i-1} \geq k_i/n$. Using now Claim 1, we conclude $ \phi(x_1,x_2,\dots,x_n) \leq \phi(x_1',x_2',\dots,x_n')$, where $\{x_1',x_2',\dots,x_n'\}=\{y_1,\dots, y_{i-1}, y_{i+1},\dots,y_j\}$, where each $y_s$ has a multiplicity $k_s$, except one of the two points $y_{i-1}$ or $y_{i+1}$. Either $y_{i-1}$ has multiplicity $k_{i-1}+k_i$ or $y_{i+1}$ has multiplicity $k_{i+1}+k_i$. Proceeding in this way, each time increasing $\phi$, we finally will get to either $(0,0,\dots,0)$ or $(1,1,\dots,1)$. Finally, \[ \max_{0\leq x_1\leq\cdots\leq x_n\leq 1}\sum_{k=1}^{n}f\left ( \left | x_k-\frac{2k-1}{2n} \right | \right )= \sum_{k=1}^{n}f\left ( \frac{2k-1}{2n} \right )\]

The answer is $$ f \left( \frac{1}{2n} \right) + f \left( \frac{3}{2n} \right) + \cdots + f \left( \frac{2n-1}{2n} \right) $$which is achieved at both $x_1 = x_2 = \cdots = x_n = 0$ or $x_1 = x_2 = \cdots = x_n =1$. Now we show its most optimal. Define $$ y_k = \left\vert x_k - \frac{2k-1}{2n} \right\vert ~~,~~ S = \{ y_1,y_2,\ldots,y_n \} = \{z_1 \le z_2 \le \cdots \le z_n \} $$Note our problem is equivalent to showing $$ z_i \le \frac{2i-1}{2n} ~~ \forall ~~ 1 \le i \le n $$Observe the above is further equivalent to following: Claim: For each $1 \le i \le n$, at least $i$ elements of set $S$ are $\le \frac{2i-1}{2n}$. Fix $i$. For each $1 \le r \le i$, define set $$ T_r = \{ y_l : l \equiv r \pmod{i} \} = \{y_r,y_{r+i},\ldots,y_{r + ki} \} \qquad \text{where } k = \left\lfloor \frac{n-r}{i} \right\rfloor $$We in fact each $T_r$ has some element $\le \frac{2i-1}{2n}$. Assume contrary and fix $r$. We will show $$ x_l > \frac{2l- 1}{2n} + \frac{2i-1}{2n} ~~ \forall ~ l = r+si, s = \in \{0,1,\ldots,k\} \qquad \qquad (\clubsuit) $$Basically, for each $l \in \{r,r+j,\ldots,r+kj\}$, we know $$ \left\vert x_l - \frac{2l-1}{2n} \right\vert > \frac{2i-1}{2n} $$so if it would further happen that $$ x_l \ge \frac{2l-1}{2n} - \frac{2j-1}{2n} \qquad \qquad (\diamondsuit)$$Then that would force $$ x_l > \frac{2l - 1}{2n} + \frac{2i-1}{2n} $$Basically $(\diamondsuit)$ implies $(\clubsuit)$. Now we induct on $s$ to prove $(\clubsuit)$. For base case $s = 0$, note $(\diamondsuit)$ is of course true for $l = r$ as $x_l \ge 0$, proving base case. Assume assertion holds for $s=t$. We prove it for $s=t+1$ now. Observe, \begin{align*} x_{r + i(t+1)} \ge x_{r + it} > \frac{2(r+it) - 1}{2n} + \frac{2i-1}{2n} = \frac{2(r + i (t+1)) - 1}{2n} - \frac{1}{2n} \ge \frac{2(r + i(t + 1)) - 1}{2n} - \frac{2j-1}{2n} \end{align*}which proves $(\diamondsuit)$, consequently implying $(\clubsuit)$. So we have finally proven $(\clubsuit)$. But observe $(\clubsuit)$ is a clear contradiction for $l = r + ki$ as then, $$ 1 \ge x_{r+ki} > \frac{2(r + ki) - 1}{2n} + \frac{2i-1}{2n} = \frac{r + i(k+1)) - 1}{n} \ge 1 $$This completes the proof. Motivation: Going in the direction of Claim is motivated by IMOSL 2003 A6. Proving more strongly that each set $T_r$ has a elements $\le \frac{2i-1}{2n}$ is motivated by the case $i=1$. Basically, we only require gaps between indices to be at most $i$ and for contradiction at end we just need $r+ki$ to be close to $n$. Obtaining $(\clubsuit)$ just comes inductively. Like for $i=1$, one gets $x_1 \ge \frac{2}{2n}$. Then $x_2 \ge \frac{4}{4n}$ and so on. Like, proving $(\clubsuit)$ for $i=1$ can be done very quickly without paper. That gave a contradiction. Generalizing to a general $i$ involved some easy paper work.

nice problem