Is every positive integer the permanent of some 0-1 matrix?

The answer to the question is yes. Given $k$, the 0-1 matrix given by

$1$ $1$ $\dotsc$ $1$ $0$ $0$ $\dotsc$ $0$ $0$ $0$ $0$

$0$ $1$ $1$ $0$ $\dotsc$ $0$ $0$ $\dotsc$ $0$ $0$ $0$

$0$ $0$ $1$ $1$ $0$ $\dotsc$ $0$ $\dotsc$ $0$ $0$ $0$

$\dotsc$

$1$ $0$ $0$ $0$ $0$ $\dotsc$ $\dotsc$ $0$ $0$ $0$ $1$

where the first row has precisely $k$ entries equal to $1$, evidently has permanent equal to $k$.

For $k=1$ the matrix is ($1$). For $k=2$ the matrix is

$1$ $1$

$1$ $1$

For $k=3$ the matrix is

$1$ $1$ $1$

$0$ $1$ $1$

$1$ $0$ $1$.

For $k=4$ the matrix is

$1$ $1$ $1$ $1$

$0$ $1$ $1$ $0$

$0$ $0$ $1$ $1$

$1$ $0$ $0$ $1$

which evidently has permanent equal to $4$.

Please note that also, for each given $k$ and $1\leq \ell \leq k$, this construction can be tweaked to give an explicit $k\times k$ sized $0$-$1$-matrix having permanent precisely $\ell$, just by making the first row have precisely $\ell$ entries equal to $1$.

Please also note that my construction does not have any bearing on the interesting and apparently difficult question which was cited in the present OP: my construction is too wasteful: it utilizes a $k\times k$ matrix, which is far too large when it comes to meet the demands of the OP in said question.

For the record, here is the construction of Kim, Lee, and Seol that I alluded to in my comment to Peter Heinig's answer.

Write $k-1$ in binary, and let $n$ be 1 plus the number of binary digits of $k-1$.

Start with an $n\times n$ matrix with all $1$'s on or above the main diagonal and all $0$'s below the diagonal. Then replace the first $n-1$ entries of the bottom row of the matrix with the binary representation of $k-1$ (one bit per entry).

For example, if $k=389$, then $k-1$ in binary is $110000100$. Then $n=1+9=10$ and the matrix is $$\matrix{ 1&1&1&1&1&1&1&1&1&1\cr 0&1&1&1&1&1&1&1&1&1\cr 0&0&1&1&1&1&1&1&1&1\cr 0&0&0&1&1&1&1&1&1&1\cr 0&0&0&0&1&1&1&1&1&1\cr 0&0&0&0&0&1&1&1&1&1\cr 0&0&0&0&0&0&1&1&1&1\cr 0&0&0&0&0&0&0&1&1&1\cr 0&0&0&0&0&0&0&0&1&1\cr 1&1&0&0&0&0&1&0&0&1\cr }$$