How do I find the total sum of the numbers occupying three different regions of a $3 \times 3$ grid?

Question

The problem is as follows:

The following board must be completed with natural numbers from $1$ to $15$ with no repetitions allowed in such a way that the sum of the numbers on each of the three squares of $3 \times 3$ is always the same. However, the sum of the numbers on each of the $3 \times 3$ squares must be the maximum as possible. What will be the result of that sum?

The options to choose from given are:

$\begin{array}{ll} 1.&87\\ 2.&86\\ 3.&89\\ 4.&88\\ \end{array}$

I've found this question in my riddles book. I've attempted to solve it but I've became tangled with different trials. Does there exist a strategy or straightforward method to solve this? Can somebody help me?

I believe what it is intended to be asked here is to attain the maximum sum in those areas which are bounded by the bold lines.

Each region is composed of nine numbers. Then I tried to select it from $1$ to $9$ as follows:

$1+2+3+4+5+6+7+8+9=45$

and the rest:

$10+11+12+13+14+15=75$

But I don't know if what I'm doing is right or not. Supposedly the answer is $86$, but I have no idea how to get there and more importantly why?.

I'd really appreciate somebody could include some drawing or sketch as part of your answer so I could better visualize how should I fill these squares or what sort of logic should be used to solve this.

Answer 1

Step 1

Since the middle four cells will be used in all of the $3 \times 3$ squares, to maximise the sum, we should put the largest numbers in here (order is irrelevant).

Step 2

Then the sum of the three cells on the left of the middle $2 \times 2$ must be the same as the sum of the three cells on the right and the sum of the three cells above must be the same as the sum of the three cells below. Once we have achieved this, we are done. These two sums involve $11$ cells with the top right-hand corner being used twice. Since the sum must be even, the top corner entry must also be even. Also, since this cell is used only once in the $3 \times 3$ squares we should pick this to be as small as possible to maximize the mutual sum, i.e, this cell should contain $2$.

Step 3

Now, we must fill in the rest. We know that the sum of the $11$ cells considered in step 2, with the top corner added twice, is $68$. To balance everything, we must therefore ensure the sum in each three-cell row or column is $17$ ($=68/4$). There are a few ways to do this but here is one.

And the mutual sum is $86$, as required.