Pyramid Scheme

Relevant wiki: Properties of Binomial Coefficients

Let's represent the bottom-most row of the pyramid by $\{a_0, a_1, a_2, \dots, a_{2016}\},$ which is a permutation of the first 2017 positive integers. Each integer represents a value in a box.

Let $S$ be the value in the top-most box. We can find that this value is

$S = \sum_{i = 0}^{2016} \binom{2016}{i}a_i = \binom{2016}{1008}a_{1008} + \sum_{i = 0}^{1007} \binom{2016}{i}(a_i + a_{2016 - i}).$

Notice that $\binom{2016}{1008} \geq \binom{2016}{1007} \geq \binom{2016}{1006} \geq \cdots \geq \binom{2016}{1} \geq \binom{2016}{0}.$ Therefore, by the Rearrangement Inequality, the maximum value of $S$ occurs when

$a_{1008} = 2017, \{a_{1007}, a_{1009}\} = \{2016, 2015\}, \{a_{1006}, a_{1010}\} = \{2014, 2013\}, \dots, \{a_{0}, a_{2016}\} = \{2, 1\},$

while the minimum value occurs when

$a_{1008} = 1, \{a_{1007}, a_{1009}\} = \{2, 3\}, \{a_{1006}, a_{1010}\} = \{4, 5\}, \dots, \{a_{0}, a_{2016}\} = \{2016, 2017\}.$

Thus,

$\begin{array}{cccccccccc} M & = & 2017\dbinom{2016}{1008} & + & (2016 + 2015)\dbinom{2016}{1007} & + & (2014 + 2013)\dbinom{2016}{1006} & + & \cdots & + & (2 + 1)\dbinom{2016}{0} \\ m & = & \dbinom{2016}{1008} & + & (2 + 3)\dbinom{2016}{1007} & + & (4 + 5)\dbinom{2016}{1006} & + & \cdots & + & (2016 + 2017)\dbinom{2016}{0}. \\ \end{array}$

If we add along each column, notice that we can pair off the coefficients $i$ and $2018 - i$ to get 2018. We have

$\begin{aligned} M + m &= 2018\dbinom{2016}{1008} + 2(2018)\dbinom{2016}{1007} + 2(2018)\dbinom{2016}{1006} + \cdots + 2(2018)\dbinom{2016}{0} \\ &= 2018 \left ( \dbinom{2016}{1008} + 2 \sum_{i = 0}^{1007} \dbinom{2016}{i} \right ) \\ &= 2018 \sum_{i = 0}^{2016} \dbinom{2016}{i} \\ &= 2018 \times 2^{2016} \\ &= 1009 \times 2^{2017}. \end{aligned}$

Finally, we have $a + b = 1009 + 2017 = \boxed{3026}.$

Generalization: For a pyramid with $k$ rows, if we fill the bottom-most row with the first $k$ positive integers, the sum of the maximum and minimum possible values of the top-most box will be $(k + 1)2^{k-1}.$

This is a rewrite of my original solution, more thoroughly reasoned per Calvin Lin's wishes.

Let $A_k$ with $k = 1, \dots, 2017$ be an arrangement of the numbers $\{1,\dots, 2017\}$ in the bottom row of the diagram. Define $A'_k = 2018 - A_k$ ; it is easy to see that $A'_k$ is also a valid arrangement of these numbers. I will call these arrangements dual to each other.

I will write $\Sigma A$ for the value in the top box generated by arrangement $A_k$ . It is easy to see that it is a weighted sum: $\Sigma A = \sum_{k = 1}^{2017} f_kA_k$ for some constants $f_k$ . We do not need the exact values of $f_k$ ; the only important thing is that their sum is $C = \sum_k f_k = 2^{2016}$ . This may be seen as follows: each box in the second row from the bottom is the sum of two terms from the bottom row; each box in the third row from the bottom is the sum of four terms from the bottom row; each box in the fourth row from the bottom, the sum of eight terms; and so on, until we reach the 2017th row from the bottom, which is the top box.

The next observation to make is that for two dual arrangements $\{A_k,A'_k\}$ , $\Sigma A + \Sigma A' = 2018C$ . Proof: $\Sigma A + \Sigma A' = \sum_{k = 1}^{2017} f_kA_k + \sum_{k = 1}^{2017} f_k(2018-A_k) = \sum_{k=1}^{2017} 2018f_k = 2018C.$

Now I claim that any arrangement with maximum sum is the dual of an arrangement with minimum sum, and vice versa. We can see that as follows. Suppose that $A_k$ is an arrangement with maximum sum, and $B'_k$ is an arrangement with minimum sum. Then $\Sigma A' = 2018C - \Sigma A \leq 2018C - \Sigma B = \Sigma B'$ (because by definition of "maximum", $\Sigma A \geq \Sigma B$ ). However, since $\Sigma B'$ is a minimum, we also have $\Sigma A' \geq \Sigma B'$ , and it follows that therefore $\Sigma A' = \Sigma B'$ is also minimal, and $\Sigma B = \Sigma A$ is also maximal.

Finally, then, choosing such a max-min pair of dual arrangements $\{A_k,A'_k\}$ , we get $M + m = \Sigma A + \Sigma A' = 2018C = 2018\cdot 2^{2016} = 1009\cdot 2^{2017},$ making the final answer $\boxed{3026}$ .

When we form one of these pyramids, the maximum value obtainable when going through the process is obtained when the middle number(s) (depending on whether the number at the bottom is even or odd) are maximized. Similarly, the minimum value is obtained when the middle number(s) is minimized. This is because as we go up the pyramid, the middle numbers expand outward and are added with the most numbers, affecting the outcome the most. Through some testing with smaller numbers, we can get a set of coordinate points, with $x$ being the largest number at the bottom and $y$ being $M+m$ : $(2, 3+3); (3, 7+9); (4, 16+24)$ . Comparing the y values to the equation given in the question, we see that the formula for any number is simply $(n+1) \cdot 2^{n-1}$ . Plugging in our values, we get $(2017+1) \cdot 2^{2017-1}=2018 \cdot 2^{2016} = 1009 \cdot 2^{2017} \implies a+b=1009+2017=\boxed{3026}$ . I know this was kind of a cheating strategy using the given equation. The complete way to go about solving this would use combinatorics, which I was too lazy to use.