Let $n \geq 1$ be an integer. How many ways can the rectangle having vertices $(0,0),(n, 0),(n, 1),(0,1)$ be dissected into $2 n$ triangles, all vertices of which have integer coordinates?

Remark. The triangles are considered as positioned on the coordinate plane; in particular, tilings related by rotation and reflection are considered distinct.
Since the vertices of the tiles must lie within the rectangle and cannot be collinear, all tiles have the form

$$
(a, 0)(b, 0)(c, 1) \text { or } \quad(a, 1)(b, 1)(c, 0) \quad(a<b) \text {. }
$$

Such a triangle has area $(b-a) / 2$; for $2 n$ such triangles to tile a rectangle of area $n$, it is necessary that they all have area $1 / 2$, i.e. be of the form

$$
(a, 0)(a+1,0)(c, 1) \quad \text { or } \quad(a, 1)(a+1,1)(c, 0) \text {. }
$$

Let $T_{a, b}$, for $a, b \geq 0$, be the trapezoid (or rectangle, triangle, degenerate segment) having vertices $(0,0),(a, 0),(b, 1),(0,1)$, and let $t_{a, b}$ be the number of ways to tile it with triangles of type (11). If $a$ and $b$ are positive, the right-hand edge can be covered in only two ways:

- Laying the tile $(a-1,0)(a, 0)(b, 1)$ and tiling the resulting $T_{a-1, b}$;
- Laying the tile $(a, 0)(b-1,1)(b, 1)$ and tiling the resulting $T_{a, b-1}$.

Thus in this case $t_{a, b}=t_{a-1, b}+t_{a, b-1}$. This, together with the initial conditions $t_{a, 0}=t_{0, b}=1$, is the recursion of Pascal's triangle; it shows that $t_{a, b}=\left(\begin{array}{c}a+b \\ a\end{array}\right)$. Thus the desired answer is $t_{n, n}=\left(\begin{array}{c}2 n \\ n\end{array}\right)$.