Picture from official booklet

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prlgr.pdf (61kb)

[asy][asy] size(170); defaultpen(linewidth(0.8)); path hex = dir(30)--dir(90)--dir(150)--dir(210)--dir(270)--dir(330)--cycle; for(int i=0;i<=3;i=i+1) { for(int j=0;j<=2;j=j+1) { real shiftx=j*sqrt(3)/2+i*sqrt(3),shifty=j*3/2; draw(shift(shiftx,shifty)*hex); } } [/asy][/asy] Merged into the original post. -mod

Juniors)The answer is k,I will post the full solution for the Junior part later

Juniors) Let's count the number of vertices in each row. Here is an example for $2 \times 3$ parallelogram. ..O...O...O..... 3 ./.\./.\./.\.... O...O...O...O... 4 |...|...|...|... O...O...O...O... 4 .\./.\./.\./.\.. ..O...O...O...O. 4 ..|...|...|...|. ..O...O...O...O. 4 ...\./.\./.\./.. ....O...O...O... 3 We can see that the topmost row and the bottommost row have $l$ vertices each, while all the other rows have $l+1$ vertices each. We will induct on $k$ that the answer is $k$ vertical segments. For $k=0$, clearly the answer is $0$. Now, note the topmost row. It doesn't have any vertical segments, so all $l$ vertices must connect to the second-topmost row with non-vertical segments. This leaves only one vertex on the second-topmost row unpaired, which must connect to the third-topmost row with a vertical segment. But now note that the remaining rows have the same structure as a $(k-1) \times l$ parallelogram: the topmost row has $l$ unpaired vertices which cannot connect vertically, the bottommost row has $l$ unpaired vertices, and the other rows have $l+1$ unpaired vertices each. Thus by induction claim, there are $k-1$ vertical segments here, and combined with the vertical segment we have just now, we have $k$ vertical segments on any valid pairing.

Senior) Let $a_{k,l}$ be the number of ways for $k \times l$ parallelogram. Then we get $a_{k,l} = a_{k-1,l} + a_{k,l-1}$. (Easy to see it dividing the case by how the upper right vertex is paired up) Also easy to see that $a_{1,k}=a_{k,1}=k+1$. By induction, we can conclude $a_{k,l} = {{k+l}\choose{\min(k,l)}}.$