Inspired by Romain Bouchard

Call a collection $A = (a_1,a_2,...,a_n)$ of distinct positive integers such that $a_1 < a_2 < \cdots < a_n$ and $a_1 + a_2 +\cdots + a_n = 2018$ a viable collection. If $A$ is a viable collection, let $P(A) = a_1a_2 \cdots a_n$ be the product of its elements. Note that $2(n-2) > n$ for $n > 4$ and that $(b-1)(a+1) > ab$ and $a+1 < b-1$ for $b > a + 2$ .

• If $A$ is a viable collection with $a_1=1$ , then the collection $\hat{A}$ obtained by removing $a_1=1$ and replacing $a_n$ with $a_n+1$ is another viable collection with $P(\hat{A}) > P(A)$ .
• If $A$ is a viable collection with $a_1 \ge 5$ , then the collection $\hat{A}$ obtained by replacing $a_1$ with $2$ and $a_1-2$ is another viable collection with $P(\hat{A}) > P(A)$ .
• If $A$ is a viable collection, and there exists $1 \le u < n$ for which $a_{u+1} > a_u + 2$ , then the collection $\hat{A}$ obtained by replacing $a_u$ and $a_{u+1}$ with $a_u+1$ and $a_{u+1}-1$ is another viable collection with $P(\hat{A}) > P(A)$ .
• If $A$ is a viable collection, and there exists $1 \le u < v < n$ for which $a_{u+1} = a_u+2$ and $a_{v+1} = a_v+2$ , then the collection $\hat{A}$ obtained by replacing $a_u$ and $a_{v+1}$ with $a_u+1$ and $a_{v+1}-1$ is another viable collection with $P(\hat{A}) > P(A)$ .

Thus, if $A$ is a viable collection with the largest possible value of $P(A)$ , then the smallest integer is $2$ , $3$ or $4$ , and the integers will be consecutive, except that at most one pair of neighbouring integers can differ by $2$ . There are three such collections:

• $2,3,...,60,62,63,64$ (start with $2$ and omit $61$ ) has product $\tfrac{64!}{61}$ .
• $3,4,...,58,60,61,62,63,64$ (start with $3$ and omit $59$ ) has product $\tfrac{64!}{2 \times 59}$ .
• $4,5,...,55,57,58,59,60,61,62,63,64$ (start with $4$ and omit $56$ ) has product $\tfrac{64!}{6 \times 56}$ .

The first of these has the largest product. This makes the answer $\tfrac12 \times 64 \times 61 = \boxed{1952}$ .