, 2026

Solution 1:

The total area of the vertical strips is $\$n\times 2\times$ 20 = $40n\text{ cm}^{2}\$.$

The total area of the horizontal strips is $\$n\times 2\times$ 26 = $52n\text{ cm}^{2}\$.$

Each horizontal strip overlaps with each vertical strip in a rectangle.
Each rectangle formed by an overlap has the same height as a horizontal
strip and the same width as a vertical strip. Therefore, each
overlapping rectangle has area $\$2\times$ 2 =
$4\text{ cm}^{2}\$.$

There are $n\times n = n^2$
overlaps in total, and each overlap is included in both the total area
of the horizontal strips and the total area of the vertical
strips.

Therefore, the total painted area is $(40n +
52n - $4n^2)\text{ cm}^{2}\$.$ We are given that this is equal to
$\dfrac{12}{13}$ of the total area of
the rectangle, or $\$\dfrac{12}{13}\times 20\times$ 26 = $480\text{ cm}^{2}\$.$

The integer $n$ satisfies the
equation $40n+52n-4n^2 = 480$, which
is equivalent to $n^2 -
23n+120=0$.

Factoring gives $(n-15)(n-8) = 0$, so
the possible values of $n$ are $n=8$ and $n=15$. However, if $n=15$, then the horizontal strips would
cover a height of $\$2\times$ 15 = 30 $\text{}$
cm}. Since the rectangle has a height of 20 $\text{}$ cm}$, and the horizontal strips
do not overlap, this is impossible.

The only possible value of $n$ is
$8$.

Solution 2:

Shifting a horizontal strip up or down, will not change the total
amount of the rectangle that is painted, as long as this shifting does
not introduce overlap between horizontal strips. Similarly, shifting
vertical strips to the left or right does not change the amount of the
rectangle that is painted, as long as the shifting does not introduce
overlap between vertical strips.

Because of this, we can assume without loss of generality
that the horizontal strips are all at the top of the rectangle and
exactly touching each other without overlap. We can also assume that the
vertical strips are all at the left of the grid touching each other
without overlap. This configuration is shown in the diagram.

[[IMAGE0]]

The unpainted portion is a rectangle with width $26-2n$ and height $20-2n$. We are also given that the area of
the painted portion is $\dfrac{12}{13}$ of the area of the $20\times26$ rectangle, so the area of the
unpainted portion must be $\dfrac{1}{13}$ of $20\times 26$.

Therefore, we have $\$(26-2n)\times(20-2n) = \dfrac{1}{13}\times 20\times$ 26=40$. Expanding the left side
gives $520 - 52n - 40n + 4n^2 = 40$,
which can be rearranged to get $4n^2-92n+480=0$.

Proceeding as in Solution 1, the only solution to this equation that
makes sense given the context is $n=8$.