Is there a positive integer $N$, besides 1 and 2, such that there is a permutation $a\_1=1,a\_2,a\_3,\dots,a\_N$ of $1,2,3,\dots,N$ in which for each $k>1$, $a\_k=a\_{k-1}\div k,a\_k=a\_{k-1}-k,a\_k=a\_{k-1}+k,\textrm{or }a\_k=a\_{k-1}\times k$?



Can one permute the positive integers with just the four arithmetical operations?
This is implicit in Gerhard Paseman's answer but it should be made explicit: what you ask for is not possible, for a fairly simple reason. Consider $a\_N$. There are four possibilities:
$$
(1)\qquad a\_N = a\_{N-1}/N
$$
but this can't be satisfied, because $a\_{N-1}\le N$, so either $a\_{N-1}$ doesn't divide $N$ or $a\_{N-1}=N$, in which case $a\_N=1$ --- but $a\_N$ can't be 1 because it was specified that $a\_1 = 1$. So we know that $a\_N \ne a\_{N-1}/N$.
$$
(2)\qquad a\_N = a\_{N-1} - N
$$
but this can't be satisfied, because $a\_{N-1}\le N$, so $a\_N$ would have to be zero or negative. So we know that $a\_N \ne a\_{N-1} - N$.
$$
(3)\qquad a\_N = a\_{N-1} + N
$$
but this can't be satisfied, because $a\_{N-1}\ge 2$, so $a\_N$ would have to be greater than $N$. So we know that $a\_N \ne a\_{N-1} + N$.
$$
(4)\qquad a\_N = N a\_{N-1}
$$
but this can't be satisfied if $N>2$, because in that case $a\_{N-1}\ge 2$, so $a\_N$ would have to be greater than $N$. So we know that $a\_N \ne N a\_{N-1}$.


So your stipulation that $a\_1=1$ makes it impossible for any $N>2$.


