Induction of the N-Block Game!

Your score is always the same no matter how you play, and $S(n) = n(n-1)/2$ . This is easy to prove by induction, since your score is $a(n-a) + S(a) + S(n-a)$ for some $a$ , and $a(n-a) + a(a-1)/2 + (n-a)(n-a-1)/2 = n(n-1)/2$ .

This is not the most satisfying solution; there is a nicer one involving inscribing squares inside isoceles right triangles (where two of the squares' sides lie on the legs of the right triangle). But I will leave this for others to find. (Your score is the total area of the squares you inscribe...)

This is IPSC 2013 Problem B . The following solution rephrases the one given in the solution booklet.

Suppose that we have $n$ blocks. Make a complete graph of $n$ vertices. Two vertices are connected iff they are in the same pile; additionally, if two vertices are connected, then they must have an edge between them. Thus the graph will be a union of disjoint cliques at all times.

Whenever one splits a pile into two, we remove the corresponding edges from the graph. For example, when one splits a pile of $5$ into $2$ and $3$ , we remove the edges from our clique of $5$ so that only cliques of $2$ and $3$ remains.

Suppose we split a pile into two piles $a,b$ . Our score increases by the product of the sizes of the two piles, namely $ab$ . However, we also remove the same number of edges from the graph; the removed edges form a complete bipartite graph $K_{a,b}$ , and there are $ab$ such edges. So we can pretend every edge removed scores $1$ point.

Now, in a complete graph of $n$ vertices, there are $\dfrac{n(n-1)}{2}$ edges. At the end, there is no edge (as no pile has more than one block, and if two vertices are connected, it means they belong to the same pile and thus a pile will have more than one block). Thus we have removed all edges, and thus we score $\dfrac{n(n-1)}{2}$ points. For $n = 2014$ , this results to $\boxed{2027091}$ .