The number of nails on the perimeter is $4n-4$ since there are 4 corners.
The total number of nails in the square is $n \times n = n^2$.
The number of nails inside the square is the total number of nails minus the number of nails on the perimeter.
Thus, the number of nails inside the square is $n^2 - (4n-4) = n^2 - 4n + 4$.
We want the number of nails on the perimeter to be equal to the number of nails inside the square, so we have
$4n - 4 = n^2 - 4n + 4$
$n^2 - 8n + 8 = 0$
Using the quadratic formula, we have
$n = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(8)}}{2(1)}$
$n = \frac{8 \pm \sqrt{64 - 32}}{2}$
$n = \frac{8 \pm \sqrt{32}}{2}$
$n = \frac{8 \pm 4\sqrt{2}}{2}$
$n = 4 \pm 2\sqrt{2}$
Since $n$ must be an integer, there is no integer value for $n$ that satisfies the condition.
However, we can rephrase the question in terms of Pick's theorem.
Let $I$ be the number of interior nails and $B$ be the number of boundary nails.
The area of the polygon is given by $A = I + \frac{B}{2} - 1$.
Since we have a square, let the side length be $s$, where $s$ is an integer. Then the area of the square is $s^2$.
Also, we are given that $I = B$.
So, $s^2 = I + \frac{I}{2} - 1 = \frac{3I}{2} - 1$.
Since $B = 4s$, we have $I = 4s$.
Thus, $s^2 = \frac{3(4s)}{2} - 1 = 6s - 1$
$s^2 - 6s + 1 = 0$
$s = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}$.
Since $s$ must be an integer, there is no square with this property.
However, if we consider the vertices to be the only nails on the perimeter, then $B = 4$. We want $I = B$, so $I = 4$.
Then $A = I + \frac{B}{2} - 1 = 4 + \frac{4}{2} - 1 = 4 + 2 - 1 = 5$.
The area of a square is $s^2$, so $s^2 = 5$, which means $s = \sqrt{5}$. Since $s$ is not an integer, this is not possible.
Final Answer: The final answer is $\boxed{no}$