Let $n$ be a positive integer. The point on the $x$-axis with abscissa $n$ is connected to the points on the $y$-axis with ordinates $n-1$ and $n+1$, and the point on the $y$-axis with ordinate $n$ is connected to the points on the $x$-axis with abscissae $n-1$ and $n+1$. What is the area of the quadrilateral enclosed by these four segments?
Let's denote the vertices of the quadrilateral as $A, B, C$, and $D$ as shown in the figure. Since $B(n ; 0)$ and $D(0 ; n)$, we have $BD = n \sqrt{2}$.

From the symmetry of the construction, it follows that points $A$ and $C$ lie on the angle bisector of the $x$ and $y$ axes, the line $y = x$, which is perpendicular to the diagonal $BD$. Thus, the area of the quadrilateral is half the product of the diagonals.

To determine the length of $AC$, we first write down the coordinates of points $A$ and $C$.

The equation of the line passing through points $D(0 ; n)$ and $E(n-1 ; 0)$ is:

$$
y = \frac{-n}{n-1} x + n
$$

The coordinates of point $A$ are given by the intersection of this line with the line $y = x$.

$$
A\left(\frac{n(n-1)}{2n-1} ; \frac{n(n-1)}{2n-1}\right)
$$

The coordinates of point $C$ are given by the intersection of the line passing through points $F(0 ; n+1)$ and $B(n ; 0)$ with the line $y = x$.

The equation of the line $FD$ is:

$$
y = \frac{n+1}{-n} x + n + 1 \quad \text{and thus} \quad C\left(\frac{n(n+1)}{2n+1} ; \frac{n(n+1)}{2n+1}\right)
$$

The distance between the two points is:

$$
AC = \sqrt{\left(\frac{n(n+1)}{2n+1} - \frac{n(n-1)}{2n-1}\right)^2 + \left(\frac{n(n+1)}{2n+1} - \frac{n(n-1)}{2n-1}\right)^2} = \sqrt{2\left(\frac{2n^2}{4n^2-1}\right)^2} = \sqrt{2} \frac{2n^2}{4n^2-1}
$$

Finally, the area of the quadrilateral is:

$$
T = \frac{BD \cdot AC}{2} = \frac{1}{2} n \sqrt{2} \cdot \sqrt{2} \frac{2n^2}{4n^2-1} = \frac{2n^3}{4n^2-1}
$$