Since $a_{ij}=0$ for $i>j$, the matrix $A$ is upper triangular.
Hence, the eigenvalues of $A$ are its diagonal entries.
For $i=j$,
$a_{ii}=\min(i+1,\,2i-1)$.
Now compute the diagonal entries:
$a_{11}=\min(2,1)=1$
$a_{22}=\min(3,3)=3$
$a_{33}=\min(4,5)=4$
$a_{44}=\min(5,7)=5$
and similarly,
$a_{55}=6,\quad a_{66}=7,\quad a_{77}=8,\quad a_{88}=9,\quad a_{99}=10,\quad a_{10,10}=11$.
So the eigenvalues are
$1,3,4,5,6,7,8,9,10,11$.
All these eigenvalues are distinct.
Therefore, $A$ has $10$ linearly independent eigenvectors.
Hence, the required answer is $\boxed{10}$.
