A man has to paint n consecutive mile posts and wants to do this as inefficiently as possible… I can't comment on this question A man has to paint n consecutive mile posts and wants to do this as inefficiently as possible... but I have further questions from this problem. 
Based on the most voted answer, 
"Now, if we wish to maximize this sum (i.e.  most inefficient), it is clear that this is maximal when $k_i = 2$ for large values of $a_i$ and $k_i = -2$ for small values of $a_i$. In particular,"
Why is the maximal when $k_i = 2$ for large values of $a_i$? If I understand correctly,  $a_i$ is the post order that he paints, so $i$ can be a small number if it is painted at the last. So how does this maximise $2i$?
Also, "I leave it to you to verify that with these coefficients, you would get a sum of $2\lfloor \frac {n}{2} \rfloor \lceil \frac {n}{2} \rceil$."
I'm a bit slow but can anybody explain how is this obtained?
 A: We want to maximize $\sum k_i a_i$ under the constraint that $(k_1,\ldots,k_n)$ belongs to an actually possible route. As a consequence of "possible route" we know that $k_i\in\{-2,0,2\}$ and $\sum k_i=0$.
So let 
$$S=\bigl\{\,(k_1,\ldots,k_n)\bigm| \forall i\colon k_i\in\{-2,0,2\}\text{ and }\textstyle\sum_ik_i=0\,\bigr\} $$
Now Calvin Lin suggest a specific such sequence $\bar{\mathbf k}=(\bar k_1,\ldots,\bar k_n)\in S$, namely 
$$ \bar k_i=\begin{cases}-2,&1\le a_i\le \lfloor n/2\rfloor\\
\hphantom{-}0,&\lfloor n/2\rfloor<a_i<\lceil(n+1)/2\rceil\\
+2,&\lceil n/2\rceil+1\le a_i\le n
\end{cases}$$
that is, first $\lfloor n/2\rfloor$ terms $=-2$, then possibly a $0$ (for  odd $n$), then $\lfloor n/2\rfloor$ terms $=+2$. More importantly, he also shows that this sequence $k_i$ indeed belongs to a possible route.
Proposition. Let $\mathbf k\in S$ be any sequence with $\mathbf k\ne \bar{\mathbf k}$. Then there exists $\mathbf k'\in S$ with $\sum k'_ia_i>\sum k_ia_i$.
Proof. As $\mathbf k\ne \bar{\mathbf k}$, there must exist an index where the two sequences differ. As $\sum k_i=\sum\bar k_i=0$, there must in fact be indices $u,v$ such that $k_u>\bar k_u$ and $k_v<\bar k_v$.
This is only possible if $a_v>\lfloor n/2\rfloor$ and $a_u<\lceil n/2\rceil+1$, at any rate $a_u<a_v$. Let $$k'_i=\begin{cases}k_i-2,&i=u\\k_i+2,&i=v\\k_i,&\text{otherwise}\end{cases} $$
Then $\mathbf k'\in S$. Also
$$\sum k'_ia_i-\sum k_ia_i=\sum (k'_i-k_i)a_i = -2a_u+2a_v=2(a_v-a_u)>0 $$
$\square$
We conclude that the optimal element of the (finite) set $S$ is $\bar{\mathbf k}$.

To compute $\sum \bar k_ia_i$, we can use that the $a_i$ are a pemutation of $1,\ldots,n$:
$$ \begin{align}\sum \bar k_ia_i&=\sum_{j=1}^{\lfloor n/2\rfloor}(-2)j+\sum_{j=\lceil n/2\rceil+1}^n2j\\&=\sum_{j=1}^{\lfloor n/2\rfloor}(-2)j+\sum_{j=1}^{\lfloor n/2\rfloor}2(j+\lceil n/2\rceil)\\
&=\sum_{j=1}^{\lfloor n/2\rfloor}2\lceil n/2\rceil\\
&=2\lfloor n/2\rfloor\lceil n/2\rceil\end{align}$$
