A game on crossing out numbers and write it as the sum On the blackboard there is the number $1$.
You can make $n$ moves, and a move consist of the following:
Erase one number on the blackboard, say $x$, and write a finite sequence of positive numbers on the blackboard, say $a_1,\ldots,a_k$, such that $\sum_{i=1}^k a_i\leq x$.
The score you get for this move is $x-\max_{i=1}^k a_i$.
As a example:
The first move I can make is remove $1$, and write numbers $1/2,1/3,1/6$ on the board. My score will be $1-1/2=1/2$. 
The second move could be remove $1/3$, and write $1/6,1/12$ on the board, since $1/6+1/12\leq 1/3$. The score for the move is $1/3-1/6=1/6$, and the total score is $1/2+1/6 = 2/3$. The board list numbers $1/2,1/6,1/6,1/12$. 
What's the maximum score possible with $n$ moves? 
A score of $\sim \frac{1}{2}\log_2 n$ can be achieved by always divide the largest number on the blackboard into two equal pieces. 
 A: I think a good strategy is to spend the first ~n/2 moves roughly splitting the largest in half, except with some adjustments so that when k ~= n/2 moves remain the sequence consists of k repeats of a single value.  Then spend the remaining moves deleting these elements one by one.
A: Only an observation, not a full answer, but too long for a comment. Let $D(x,n)$ denote the score resulting from starting from the single number $x$ and doing the "Divide the Largest Remaining Number in Two" strategy described in the question. As pointed out, $D(1,n) \sim (\log n)/2\log 2$; but it's clear that $D(x,n) = xD(1,n)$. Note also that $D(1,n) = \frac12 + D(\frac12,\frac{n-1}2) + D(\frac12,\frac{n-1}2)$: on the first move we split $\{1\}$ into $\{\frac12,\frac12\}$, then allocate half the remaining moves to doing the DLRNT strategy on the first remaining $\frac12$, and half the remaining moves to doing the DLRNT strategy on the second remaining $\frac12$.
But we could also split $\{1\}$ into $\{x,1-x\}$ on the first move (WLOG, $x\le\frac12$), then allocate a proportion $\alpha$ of the remaining moves to doing the DLRNT strategy on the remaining $x$, and a proportion $1-\alpha$ of the remaining moves to doing the DLRNT strategy on the remaining $1-x$. This yields a score of
\begin{align*}
x + D(x,\alpha(n-1)) & + D(1-x,(1-\alpha)(n-1)) \\
&\sim x + x \frac{\log[\alpha(n-1)]}{2\log 2} + (1-x) \frac{\log[(1-\alpha)(n-1)]}{2\log 2} \\
&= x + \frac{\log(n-1)}{2\log 2} + \frac{x \log \alpha + (1-x)\log(1-\alpha)}{2\log 2}.
\end{align*}
For any choice of $x$, this expression is maximized at $\alpha=x$, yielding a score of
$$
\frac{\log(n-1)}{2\log 2} + x + \frac{x \log x+ (1-x)\log(1-x)}{2\log 2}
$$
This function of $x$ equals $0$ at $x= \frac12$, and is negative for $0<x<\frac12$; but it also approaches $0$ as $x\to0^+$. In other words, using the first move to split $\{1\}$ into $\{0.001,0.999\}$, then using $99.9\%$ of the remaining moves to do the DLRNT strategy on $0.999$ and $0.1\%$ of the remaining moves to do the DLRNT strategy on $0.001$, does almost as well as the true DLRNT strategy.
Anyway, perhaps that suggests that the DLRNT strategy is asymptotically optimal. (Strategies such as Divide the Largest Remaining Number in $k$ Equal Parts do demonstrably worse as $k$ increases.)
