The sides of an equilateral triangle with sides of length $n$ have been divided into equal parts, each of length $1$, and lines have been drawn through the points of division parallel to the sides of the triangle, thus dividing the large triangle into many small triangles. Nils has a pile of rhombic tiles, each of side $1$ and angles $60^\circ$ and $120^\circ$, and wants to tile most of the triangle using these, so that each tile covers two small triangles with no overlap.  In the picture, three tiles are placed somewhat arbitrarily as an illustration. How many tiles can Nils fit inside the triangle?
[asy]
/* original code by fedja: https://artofproblemsolving.com/community/c68h207503p1220868
modified by Klaus-Anton: https://artofproblemsolving.com/community/c2083h3267391_draw_me_a_grid_of_regular_triangles
*/
size(5cm);
int n=6;
pair A=(1,0), B=dir(60);
path P=A--B--(0,0)--cycle;
path Pp=A--shift(A)*B--B--cycle;

/*
label("$A$",A,S);
label("$B$",B,dir(120));
label("$(0,0)$",(0,0),dir(210));

fill(shift(2*A-1+2*B-1)*P,yellow+white);
fill(shift(2*A-1+2*B-0)*P,yellow+white);
fill(shift(2*A-1+2*B+1)*P,yellow+white);
fill(shift(2*A-1+2*B+2)*P,yellow+white);

fill(shift(1*A-1+1*B)*P,blue+white);
fill(shift(2*A-1+1*B)*P,blue+white);
fill(shift(3*A-1+1*B)*P,blue+white);
fill(shift(4*A-1+1*B)*P,blue+white);
fill(shift(5*A-1+1*B)*P,blue+white);

fill(shift(0*A+0*B)*P,green+white);
fill(shift(0*A+1+0*B)*P,green+white);
fill(shift(0*A+2+0*B)*P,green+white);
fill(shift(0*A+3+0*B)*P,green+white);
fill(shift(0*A+4+0*B)*P,green+white);
fill(shift(0*A+5+0*B)*P,green+white);

fill(shift(2*A-1+3*B-1)*P,magenta+white);
fill(shift(3*A-1+3*B-1)*P,magenta+white);
fill(shift(4*A-1+3*B-1)*P,magenta+white);

fill(shift(5*A+5*B-5)*P,heavyred+white);

fill(shift(4*A+4*B-4)*P,palered+white);
fill(shift(4*A+4*B-3)*P,palered+white);


fill(shift(0*A+0*B)*Pp,gray);
fill(shift(0*A+1+0*B)*Pp,gray);
fill(shift(0*A+2+0*B)*Pp,gray);
fill(shift(0*A+3+0*B)*Pp,gray);
fill(shift(0*A+4+0*B)*Pp,gray);

fill(shift(1*A+1*B-1)*Pp,lightgray);
fill(shift(1*A+1*B-0)*Pp,lightgray);
fill(shift(1*A+1*B+1)*Pp,lightgray);
fill(shift(1*A+1*B+2)*Pp,lightgray);

fill(shift(2*A+2*B-2)*Pp,red);
fill(shift(2*A+2*B-1)*Pp,red);
fill(shift(2*A+2*B-0)*Pp,red);

fill(shift(3*A+3*B-2)*Pp,blue);
fill(shift(3*A+3*B-3)*Pp,blue);

fill(shift(4*A+4*B-4)*Pp,cyan);

fill(shift(0*A+1+0*B)*Pp,gray);
fill(shift(0*A+2+0*B)*Pp,gray);
fill(shift(0*A+3+0*B)*Pp,gray);
fill(shift(0*A+4+0*B)*Pp,gray);
*/

fill(Pp, rgb(244, 215, 158));
fill(shift(dir(60))*P, rgb(244, 215, 158));

fill(shift(1.5,(-sqrt(3)/2))*shift(2*dir(60))*Pp, rgb(244, 215, 158));
fill(shift(1.5,(-sqrt(3)/2))*shift(2*dir(60))*P, rgb(244, 215, 158));

fill(shift(-.5,(-sqrt(3)/2))*shift(4*dir(60))*Pp, rgb(244, 215, 158));
fill(shift(.5,(-sqrt(3)/2))*shift(4*dir(60))*P, rgb(244, 215, 158));

for(int i=0;i<n;++i){
for(int j;j<n-i;++j)
{draw(shift(i*A+j*B)*P);}}

shipout(bbox(2mm,Fill(white)));
[/asy]
1. **Understanding the Problem:**
   - We have an equilateral triangle with side length \( n \).
   - The triangle is divided into smaller equilateral triangles of side length 1.
   - We need to tile the large triangle using rhombic tiles, each of which covers two small triangles.
   - We aim to determine the maximum number of such tiles that can fit inside the large triangle.

2. **Total Number of Small Triangles:**
   - The number of small triangles in an equilateral triangle of side length \( n \) is given by the sum of the first \( n \) natural numbers:
     \[
     \text{Total small triangles} = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}
     \]

3. **Number of Rhombic Tiles:**
   - Each rhombic tile covers 2 small triangles.
   - Therefore, the maximum number of rhombic tiles that can fit is:
     \[
     \text{Maximum number of tiles} = \left\lfloor \frac{\frac{n(n+1)}{2}}{2} \right\rfloor = \left\lfloor \frac{n(n+1)}{4} \right\rfloor
     \]
   - However, we need to show that the exact number of tiles that can fit is \( \frac{n^2 - n}{2} \).

4. **Inductive Proof:**
   - **Base Case:**
     - For \( n = 2 \):
       - The triangle has \( \frac{2(2+1)}{2} = 3 \) small triangles.
       - The maximum number of rhombic tiles is \( \left\lfloor \frac{3}{2} \right\rfloor = 1 \).
       - This matches \( \frac{2^2 - 2}{2} = 1 \).

   - **Inductive Step:**
     - Assume the result is true for \( n = k \), i.e., the maximum number of tiles is \( \frac{k^2 - k}{2} \).
     - Consider the case for \( n = k+1 \):
       - The total number of small triangles is \( \frac{(k+1)(k+2)}{2} \).
       - The number of rhombic tiles is \( \left\lfloor \frac{(k+1)(k+2)}{4} \right\rfloor \).
       - We need to show that this equals \( \frac{(k+1)^2 - (k+1)}{2} \).

5. **Verification:**
   - The number of small triangles in the \( k+1 \) triangle is:
     \[
     \frac{(k+1)(k+2)}{2}
     \]
   - The number of rhombic tiles is:
     \[
     \left\lfloor \frac{(k+1)(k+2)}{4} \right\rfloor
     \]
   - Simplifying \( \frac{(k+1)^2 - (k+1)}{2} \):
     \[
     \frac{(k+1)^2 - (k+1)}{2} = \frac{k^2 + 2k + 1 - k - 1}{2} = \frac{k^2 + k}{2} = \frac{k(k+1)}{2}
     \]
   - This matches the inductive hypothesis, confirming the result.

The final answer is \( \boxed{\frac{n^2 - n}{2}} \).