Yes. Allow me to transpose your set-up, for reasons of my own dyslexia, and instead consider an $m \times n$ matrix $M$ with $m < n$. This matrix will have
block decomposition $M = \big( D \, \big| \, A \big)$ where
$\bullet$ $D$ is the $m \times m$ anti-diagonal matrix
whose $k$-th row entry is $(-1)^{k+1}$
$\bullet$ $A$ is an $m \times (n-m)$ matrix (to be specified later)
Each $m \times m$ submatrix of $M$ is uniquely prescribed by its columns, which we index
by an $m$-element subset
$I = \big\{i_1 < \cdots < i_m \big\} \subset
\big\{ 1, \dots , n \big\}$. The determinant
of the corresponding $m \times m$ submatrix
will be denoted $[I]$. By design $[I]$ equals
the determinant of the submatrix of $A$ whose
row set $\mathrm{p}(I)$ and column set $\mathrm{q}(I)$ are
\begin{equation}
\begin{array}{ll}
\mathrm{p}(I) \ &= \ \big\{ m - j +1 \ \big| \
j \in \{1, \dots, m \} \, - \, I \big\} \\
\mathrm{q}(I) \ &= \ \big\{ j - m \ \big| \ j \in I \, -\, \{1, \dots, m\} \big\}
\end{array}
\end{equation}
It will be convenient to represent an $m$-element
subset of $\{1, \dots, n \}$ by an integer partition
$\lambda = \big(\lambda_1 \geq \cdots \geq \lambda_m \big)$ with at most $m$ non-zero parts and
with $\lambda_1 \leq n - m$. The operative direction of this one-to-one correspondence assigns a partition $\lambda = \big( \lambda_1 \geq \cdots \geq \lambda_m \big)$ to the $m$-element subsets
$I^\lambda := \big\{ 1 + \lambda_m < \cdots <
m + \lambda_1 \big\}$. For brevity's sake, I shall
write $[\lambda]$ in place of $[I^\lambda]$.
Under this bijection, the matrix entry $A_{i,j}$ of $A$ correspond to the hook partition
\begin{equation}
\langle i , j \rangle \ := \
\big(j \geq
\underbrace{1 \geq \cdots \geq 1}_{\text{$i-1$ many}} \big)
\end{equation}
Given any partition $\lambda$ let $s_\lambda ( \Bbb{x} )$ denote the associated Schur function in infinitely many variables $\Bbb{x} = \big(x_1, x_2, x_3, \dots \big)$. Let us now specify the matrix entries of $A$
by setting
\begin{equation}
A_{i,j} \ := \ s_{\langle i, j \rangle}(\Bbb{x})
\end{equation}
then the so-called Giambelli identity asserts that
\begin{equation}
s_\lambda (\Bbb{x}) \ = \ [\lambda]
\end{equation}
Furthermore if we select a integer threshold $N \geq m$
and perform the specialization $x_i = 1$ whenever
$1 \leq i \leq N$ and $x_i = 0$ for $i > N$ then
\begin{equation}
s_\lambda \big(\underbrace{1, \dots, 1}_{\text{$N$ times}} , 0, 0, 0, \dots \big) \ = \ \dim \Bbb{S}_\lambda \big( \Bbb{C}^N \big)
\end{equation}
where $\Bbb{S}_\lambda \big( \Bbb{C}^N \big)$
is the irreducible representation of $\mathrm{SL}_\mathrm{N}(\Bbb{C})$ associated to
the partition $\lambda$.
This value, which can be calculated using the content formula, is a positive integer for each
$\lambda$ with at most $N$ non-zero parts.
In particular the determinants of all maximal $m \times m$ submatrices of $M$ will be positive integers.