This problem is the general version of Problem 16 and the first part of the solution is the same. Let $x$ be the side length of the square tile. If the rectangle is completely tiled with such square tiles, there exist integers $a, b$ such that the side lengths of the rectangle are $ax$ and $bx$. The diagonals of the rectangle are diameters of the circle and so their length is $2n$. Applying Pythagoras to the right angled triangle obtained by cutting the rectangle along one of its diagonals, we obtain
$$(2n)^2 = x^2(a^2 + b^2).$$

As $x, n, a, b$ are integers, this implies that $x$ is a factor of $2n$ and so $c = 2n/x$ is an integer that satisfies $c^2 = a^2 + b^2$. Because we are to find the largest possible $x = 2n/c$, we are interested in the smallest $c$ that divides $2n$ and which appears as the hypotenuse of a right angled triangle with integer side lengths.
If we have a divisor $c$ of $2n$ and a Pythagorean Triple $(a, b, c)$ with $c^2 = a^2+b^2$, then $x = 2n/c$ is an integer which can be used as the length of a tile with which we can tile an inscribed rectangle of side lengths $ax$ and $bx$.
If the Pythagorean Triple $(a, b, c)$ is not primitive, there exists an integer $t$ and a primitive Pythagorean Triple $(a', b', c')$ such that $(a, b, c) = t(a', b', c')$. Because $c = tc'$ divides $2n$, $c'$ divides $2n$ as well. Since $c' \le c$, to find the smallest possible $c$ it is sufficient to consider primitive Pythagorean Triples.
It is well known that, up to interchanging $a$ and $b$, all primitive Pythagorean Triples $(a, b, c)$ can be obtained as follows from integers $u > v$ that are coprime and for which $uv$ is even:
$$a = u^2 - v^2 \quad b = 2uv \quad c = u^2 + v^2.$$
Therefore, we are interested in finding the smallest positive divisor of $2n$ which can be written as the sum of two distinct and coprime squares of positive integers. Even though it is well known which integers can be written as a sum of two squares, we do not suppose the reader to be familiar with this theory.
If $c = u^2 + v^2$ is divisible by a prime $p$ which is congruent to 3 modulo 4, then both, $u$ and $v$, must be divisible by $p$, because otherwise there would exist an integer $s$ for which $s^2 \equiv -1 \pmod p$, which is impossible by Fermat's Little Theorem. Therefore, the desired smallest $c$ cannot be divisible by such a prime number.
It is also well known that an odd prime number can be written as the sum of two squares if and only if it is congruent to 1 modulo 4. If a divisor $c = u^2 + v^2$ of $2n$ is divisible by such a prime $p$, then $p \le c$ and $c$ can only be the smallest possible choice if $c = p$.
If $c$ is not divisible by any odd prime then $c = 2^k$. The only way to write $2^k$ as the sum of two squares is $2^{2m} = (2^m)^2$ when $k = 2m$ is even, and $2^{2m+1} = (2^m)^2 + (2^m)^2$ when $k = 2m + 1$ is odd. To see this, divide the equation $2^k = u^2 + v^2$ across by the highest possible power of 2 which leads either to $1 = 1+0$ or to $2 = 1+1$, because at least one of the three numbers must be odd. Hence, such numbers $c$ are not of the required form $u^2 + v^2$ with $u > v > 0$.
Hence, the desired smallest $c$ that divides $2n$ and which appears as the hypotenuse of a right angled triangle with integer side lengths is the smallest prime factor $p$ of $n$ which is congruent to 1 modulo 4. If no such prime factor exists, the desired tiling is not possible. If such $p$ exists, then $x = 2n/p$ is the maximal side length of a tile.
If $p = u^2 + v^2$ with $u > v > 0$, then $u, v$ are automatically coprime and $uv$ is even. We then obtain $a = u^2 - v^2$ and $b = 2uv$ and so the side lengths of the inscribed rectangle are equal to
$$ax = \frac{2an}{p} = \frac{2(u^2 - v^2)n}{u^2 + v^2} \quad \text{and} \quad bx = \frac{2bn}{p} = \frac{4uvn}{u^2 + v^2}.$$