When both sides of the triangle are divided into an equal number of steps 
(let's call this number -- n), the number of triangles is n^3 (n in the 
third power). For the example you give in your page n=3 and the answer is 
3^3 = 27.

When sides are subdivided into a different number of subdivisions, say, n 
and m, the number of triangles is equal to

    1
   --- m * n (m + n)
    2

which is integer for any integer m and n.

In J language (see http://www.jsoftware.com/ for description and download) 
the first formula is coded and invoked as

    nt =: ^&3
    nt 3
27

The second formula is coded and invoked as

    nt =: *-:@*+
    3 nt 3
27
    2 nt 5
35

The first variant of the program is three characters, the second is 6 
characters.

It took me 15-20 minutes of drawing rectangles to derive the formula.

J is an array-oriented language, descendent of APL. Therefore, the above 
programs (without change) can indeed be used to process millions of 
rectangles very fast. In order to achieve this the arguments must be 
arrays (of equal length in the second case). For example:

    (3 2) nt (3 5)
27 35


How the formula was developed:

I've put (in the thread of livejournal I referenced, 
http://www.livejournal.com/users/dr_klm/51584.html?thread=435072#t435072) 
a scan of the page, where I drew triangles during those "15-20 minutes". 
The direct link is here: http://galaxy.fzu.cz/~metlov/Triangles_Deriv.gif

I do not know it that will be enough to communicate the basic idea used 
for deriving the formula. On the other hand I do not have time to explain 
it to the arbitrary level of detalization.

The interesting part of the scan occupies the lower left quarter of the 
page. Triangles are counted separately for two lower corners of the big 
triangle (left and right) and then the result is multiplied by 2 (if n=m), 
or added up with exchanging n<->m (if n!=m). To count triangles for one 
corner I sum up the triangles, occupying all single sub-sectors, the 
triangles, occupying all pairs of two consecutive sub-sectors, ... three 
sub-sectors... etc... In this sum, the triangles, which include both left 
and right corners of the big triangle are counted with weight 1/2 (to note 
that they will be counted again, in the sum for the other corner).

I run this procedure for n=3, m=3 approximately in the middle of the 
scanned page. Then, by induction, write a general formula with the sum. 
The sum is nothing else but an arithmetic progression, which is 
immediately summed up. Then, with very basic algebra the final formulas 
are obtained.