Ion Express

We do not need to handle the requested energies individually. It is enough to construct one assignment that realizes every integer from $1$ to $N^2$. Such an assignment always exists for $N \ge 5$, so the answer is always YES.

Every route makes $N-1$ right moves and $N-1$ down moves. Let $x_i$ be the number of right moves made in row $i$. Then

$$x_i \ge 0, \qquad \sum_{i=1}^{N}x_i=N-1.$$

Conversely, every integer sequence satisfying these conditions defines a route: make $x_1$ right moves in the first row, move down, make $x_2$ right moves in the second row, and continue in this way.

Thus, by using uniform station values and horizontal signs within each row, we can express a route's energy in terms of its horizontal move counts.

1. Station values and horizontal rails

Assign the station values as follows. The starting value $c$ will be determined later.

• In the first row, assign $c$ to the starting station and $1$ to every other station.
• Assign $2$ to every station in the second row.
• Assign $i-1$ to every station in row $i$ ($3 \le i \le N$).

Make every horizontal rail in the first two rows -, and every other horizontal rail +.

Depending on its row, a right move contributes one of

$$-1,-2,2,3,\cdots,N-1$$

to the energy. Define

$$W=\{-2,-1,2,3,\cdots,N-1\}.$$

For each boundary between consecutive rows, use the same vertical sign in every column. Every route crosses each row boundary exactly once. Therefore, the combined contribution $B$ of the starting station and all down moves is independent of the route. We will arrange that $B=2N-1$.

A route's energy will then be $2N-1$ plus the sum of $N-1$ elements of $W$, with repetitions allowed. Any chosen multiset can be arranged in row order to obtain an actual route, so traversal order imposes no additional restriction.

2. Proof that every required energy is achievable

Let $kW$ denote the set of sums of $k$ elements of $W$, allowing repetitions. We prove

$$kW=[-2k,(N-1)k] \qquad (k \ge 3),$$

where an interval denotes all integers between its endpoints, inclusive.

First consider $k=3$. Classify the three chosen elements by the number of positive elements:

• No positive elements: $[-6,-3]$.
• One positive element: $[-2,N-3]$.
• Two positive elements: $[2,2N-3]$.

The negative choices are consecutive integers $-2,-1$, and the positive choices form the integer interval from $2$ to $N-1$. Consequently, every integer in each listed interval is achievable. For $N \ge 5$, these intervals join without any gaps, giving

$$3W=[-6,3N-3].$$

Now suppose that $kW=[-2k,(N-1)k]$. Appending either $-2$ or $N-1$ produces every integer in the intervals

$$[-2k-2,(N-1)k-2],$$ $$[-2k+N-1,(N-1)(k+1)],$$

respectively. There is no gap between them: the start of the second interval minus the end of the first is

$$N+1-k(N+1)=(1-k)(N+1) \le 0.$$

Therefore, $(k+1)W=[-2(k+1),(N-1)(k+1)]$.

Since $N \ge 5$, we have $N-1 \ge 3$. The achievable route energies therefore include every integer in

$$[2N-1-2(N-1),\;2N-1+(N-1)^2]=[1,N^2].$$

Hence all requested energies are satisfied simultaneously.

3. Adjusting the vertical contribution

It remains to choose the vertical signs and the starting value $c$ so that $B=2N-1$.

The station values entered by down moves, from top to bottom, are

$$2,2,3,4,\cdots,N-1.$$

If their signed sum is $S$, then $B=c+S$.

Set $G=2N-3$ and start with $S=0$. Process these values in reverse order:

$$N-1,N-2,\cdots,2,2.$$

For the current value $w$, choose the sign as follows:

• If $S<G$, make every vertical rail at this row boundary + and add $w$ to $S$.
• Otherwise, make every vertical rail at this row boundary - and subtract $w$ from $S$.

Finally, set

$$c=2N-1-S.$$

This ensures $B=2N-1$.

We must also show that $c$ is a legal station value. Write $d=G-S$. Each step updates it by

$$d \leftarrow \begin{cases}d-w & d>0,\\ d+w & d \le 0.\end{cases}$$

Initially, $d=2N-3$. After processing $N-1$ and then $N-2$, it becomes $N-2$ and then .

Next process $w=N-3,N-4,\cdots,2$. If $|d| \le w+1$ before a step, then afterwards. Thus, after these steps, .

Processing the final remaining $2$ leaves $d$ in $\{-1,0,1,2\}$. Hence

$$c=2N-1-S=d+2 \in \{1,2,3,4\}.$$

Since $N \ge 5$, the starting value is always between $1$ and $N$. Every other station value is also within the allowed range.

Solution Written by GPT6