For *n*>2, we will show a winning arrangement and a winning strategy
for the second player. This strategy revolves around the concept of "traps".
A trap is an odd-sized connected set of high-valued cells surrounded by
low-valued cells in the set of all its immediate neighbors, which must also be
of odd size.

Here is an example of such a trap, with diamonds signifying high values and empty cells signifying low values.

Notably, the total number of cells composing a trap is even.

Suppose that we tile the board into connected sets, each of even size. A feasible strategy of the second player will be in this case "Consider only values that are of the same set as the first player took from in her latest turn. Of these, remove the highest possible." If one of the sets is a trap and the first player steps into it (removes a first value from it), this cell will necessarily contain a low value. The second player, following her strategy, will take a high value. Ultimately, the trap will be emptied with the second player taking at least one more high value than the first.

We will need an arrangement with at least three traps. The reason for this is that the first player can choose where to begin in the first move, and she can avoid one trap by beginning in one of its high-valued positions. As the first player can choose to do this with the trap for which the prize is highest, we will need at least two additional traps to counter-balance this.

We will not use the kidney-bean shaped trap described above, because it is too large for our purposes. Instead, we will use the smallest possible trap:

This is a trap, provided that the middle piece of the trap is an edge piece of the board, and therefore has exactly three neighbors. On a 4x4 board, four such traps can be placed in the following arrangement:

On any larger board, we'll place four such traps along with domino tiles to complete the rest of the board in a formation such as the one demonstrated below for 8x8:

We will place in each trap the values *a*, *a*+1 and *a*+2
for the low-valued cells and *b* for the high-valued cell, for some
such *a* and *b*. This configuration insures that the sum of values
taken by the second player is at least *b*-*a*-3 larger than that
of the first player, for cells taken from the trap, provided that the first
player doesn't choose to begin the game in the high-valued cell.
If she does, the sum of the cell values taken from the trap by
the first player will be no more than *b*-*a*-1 greater than that
of the second player.

We will place in each domino piece two consecutive numbers, insuring that the advantage gained by the first player on each such tile is at most 1.

Suppose the values used in the four traps are:

The best strategy of the first player is to begin the game inside the first
trap. This will give her *n*^{2}-5 for that trap. She will also
get one point for each of the *n*^{2}/2-8 domino tiles. However,
she will lose *n*^{2}-9 for the second trap,
*n*^{2}-11 for the third and *n*^{2}-13 for the
fourth, resulting in a total balance of 1.5*n*^{2}-20, which,
for *n*≥4 is in favor of the second player.

The nice thing about this arrangement is that both players have a
clear best strategy: the first player starts with the high-valued
cell of the trap with the highest *b*-*a* value, then continues
by always picking the highest possible value, *excluding* values
within the traps. Once all the rest of the board is empty, the first player
goes into the traps in such a way that the *a*+2 and *a*+1 values
will be hers.

The strategy for the second player (given the strategy of the first player and the arrangement of the board) is as follows: if the first player took a value from a trap, take the highest possible value within that trap. If not, take the highest possible value from outside any trap.

Now, the question is merely what the values of the traps are, to determine who wins. Here is a possible arrangement (the notation of which will be explained immediately aftrwards):

ZZZAAABBBCCCDDD .... XXXYYYYX .... DCBAZ

The notation used here is as follows: the first letter signifies the
identity of the trap to which the value *n*^{2}-4*k*+1
belongs. The next, signifies the identity of the trap to which the value
*n*^{2}-4*k*+2 belongs, and so on, with the last letter
signifying the identity of the trap to which the value *n*^{2}
belongs. The first three values and the last value
belong to the trap denoted as "Z". This means that this trap has
*n*^{2}-4*k*+1 as its *a* value and
*n*^{2} as its *b* value, making it the highest valued
trap.

In the arrangment, the second
player can reach a tie on the "Y" trap, gain 4 points on the "X" trap,
and so on, with the "A" trap allowing the second player to gain 4*k*-8
points on it. In both cases, the "Z" trap can, potentially, be used by the
second player to gain 4*k*-4 points, but, in fact, the first player will
start within it, taking the *n*^{2} value first, so it will
actually give the first player an advantage of 4*k*-2.

Consider the final total balance.
The first player can gain (*n*^{2}-4*k*)/2 points for all
values outside traps plus an additional 4*k*-2 on the "Z" trap, giving
her a total advantage of *n*^{2}/2+2*k*-2. However,
the remaining traps set the balance back by
2(*k*-2)(*k*-1) to a total of
*n*^{2}/2-2*k*^{2}+8*k*-6 or
*n*^{2}/2-2(*k*-2)^{2}+2.

Choosing *k*
to be *n*/2+2 makes the first player win by exactly two points.

Now, we will tweak this arrangement slightly, by replacing the high values of the "Z" trap and the "A" trap to form this new arrangement:

ZZZAAABBBCCCDDD .... XXXYYYYX .... DCBZA

The "Z" trap, though its *b* value has been lowered by 1, is still the
most valuable trap, so the first player will start from it. On the other hand,
the "A" trap's *b* value has been raised by 1. This sets the first
player's total balance back by 2, causing a tie exactly.

The only remaining question is whether *k*=*n*/2+2 can be afforded
in every case where *n*>2. In fact, one can arrange along the sides
of an *n* by *n* board a number of traps that is equal to
4*n*/3 up to a constant. By simple trial and error, one can find that
a 4x4 board accomodates four traps, a 6x6 - five and an 8x8 - eight. (One
would have been able to fit six on a 6x6 board, if we didn't require for the
remaining pieces to be connected.) For any larger shape, note that one can
create an arrangement for an (*n*+6)x(*n*+6) board given an
*n* by *n* board by simply keeping each trap at a constant position
relative to its nearest corner and adding two traps to each side in the
positions that become freed. This adds 8 traps when raising *n* by six,
which is much more than the required three.