Solution:

The answer is in the affirmative for all $n \geq 2$. Induct on $n$. Leaving aside the trivial case $n=2$, deal first with particular cases $n=5$ and $n=6$.

If $n=5$, remove first the pair $(2,5)$, notice that $3=|2-5|$ is already on the board, so $7=2+5$ alone is written down. Removal of the pair $(3,4)$ then leaves exactly two numbers on the board, $1$ and $7$, since $|3 \pm 4|$ are both already there.

If $n=6$, remove first the pair $(1,6)$, notice that $5=|1-6|$ is already on the board, so $7=1+6$ alone is written down. Next, remove the pair $(2,5)$ and notice that $|2 \pm 5|$ are both already on the board, so no new number is written down. Finally, removal of the pair $(3,4)$ provides a single number to be written down, $1=|3-4|$, since $7=3+4$ is already on the board. At this stage, the process comes to an end: $1$ and $7$ are the two numbers left.

In the remaining cases, the problem for $n$ is brought down to the corresponding problem for $\lceil n / 2\rceil < n$ by a finite number of moves. The conclusion then follows by induction.

Let $n=4k$ or $4k-1$, where $k$ is a positive integer. Remove the pairs $(1,4k-1), (3,4k-3), \ldots, (2k-1,2k+1)$ in turn. Each time, two odd numbers are removed, and the corresponding $c=|a \pm b|$ are even numbers in the range $2$ through $4k$, of which one is always $4k$. These even numbers are already on the board at each stage, so no $c$ is to be written down, unless $n=4k-1$ in which case $4k$ is written down during the first move. The outcome of this $k$-move round is the string of even numbers $2$ through $4k$ written down on the board. At this stage, the problem is clearly brought down to the case where the numbers on the board are $1,2, \ldots, 2k=\lceil n / 2\rceil$, as desired.

Finally, let $n=4k+1$ or $4k+2$, where $k \geq 2$. Remove first the pair $(4,2k+1)$ and notice that no new number is to be written down on the board, since $4+(2k+1)=2k+5 \leq 4k+1 \leq n$. Next, remove the pairs $(1,4k+1), (3,4k-1), \ldots, (2k-1,2k+3)$ in turn. As before, at each of these stages, two odd numbers are removed; the corresponding $c=|a \pm b|$ are even numbers, this time in the range $4$ through $4k+2$, of which one is always $4k+2$; and no new numbers are to be written down on the board, except $4=|(2k-1)-(2k+3)|$ during the last move, and, possibly, $4k+2=1+(4k+1)$ during the first move if $n=4k+1$. Notice that $2$ has not yet been involved in the process, to conclude that the outcome of this $(k+1)$-move round is the string of even numbers $2$ through $4k+2$ written down on the board. At this stage, the problem is clearly brought down to the case where the numbers on the board are $1,2, \ldots, 2k+1=\lceil n / 2\rceil$, as desired.

Solution:

We will prove the following, more general statement:

Claim. Write down a finite number (at least two) of pairwise distinct positive integers on a board. A move consists in erasing any two numbers $a$ and $b$, and, for each $c$ in $\{a+b,|a-b|\}$, writing $c$ down on the board, unless $c$ is already there; if $c$ is already on the board, do nothing. Then it is possible to be left with exactly two numbers on the board after a finite number of moves.

Notice that, if we divide all numbers on the board by some common factor, the resulting process goes on equally well. Such a reduction can therefore be performed after any move.

Notice that we cannot be left with less than two numbers. So it suffices to show that, given $k$ positive integers on the board, $k \geq 3$, we can always decrease their number by at least $1$. Arguing indirectly, choose a set of $k \geq 3$ positive integers $S=\{a_1, \ldots, a_k\}$ which cannot be reduced in size by a sequence of moves, having a minimal possible sum $\sigma$. So, in any sequence of moves applied to $S$, two numbers are erased and exactly two numbers appear on each move. Moreover, the sum of any resulting set of $k$ numbers is at least $\sigma$.

Notice that, given two numbers $a > b$ on the board, we can replace them by $a+b$ and $a-b$, and then, performing a move on the two new numbers, by $(a+b)+(a-b)=2a$ and $(a+b)-(a-b)=2b$. So we can double any two numbers on the board.

We now show that, if the board contains two even numbers $a$ and $b$, we can divide them both by $2$, while keeping the other numbers unchanged. If $k$ is even, split the other numbers into pairs to multiply each pair by $2$; then clear out the common factor $2$. If $k$ is odd, split all numbers but $a$ into pairs to multiply each by $2$; then do the same for all numbers but $b$; finally, clear out the common factor $4$.

Back to the problem, if two of the numbers $a_1, \ldots, a_k$ are even, reduce them both by $2$ to get a set with a smaller sum, which is impossible. Otherwise, two numbers, say, $a_1 < a_2$, are odd, and we may replace them by the two even numbers $a_1+a_2$ and $a_2-a_1$, and then by $\frac{1}{2}(a_1+a_2)$ and $\frac{1}{2}(a_2-a_1)$, to get a set with a smaller sum, which is again impossible.