If you look at dots in the top row, you will see that they are put together into groups of 1, 2, 3, 4, and so on.

You need 6 tests:
(1, 2) → (3, 4) → (5, 6) → (7, 8) → (7, 9) → (8, 9)
If at least one of these tests is positive, then you have found two working lanterns.
It all of these tests are negative, then lantern #10 must be working. Indeed, since at least one lantern in each of the pairs (1, 2), (3, 4), (5, 6) is not working. Therefore, there are at least 2 working lanterns among #7, #8, #9, #10. If #10 is not working, then at least one of the pairs (7, 8), (7, 9), or (8, 9) must yield a positive test, which is a contradiction.

Notice that the number of small squares the main diagonal passes through is equal to the number of horizontal and vertical lines it intersects. Indeed, every time the diagonal goes through the interior of one square to the interior of another, it must intersect one of these lines.
There are 1000 + 1004 = 2004 lines which are intersected by the main diagonal. However, on four occasions (which is the greatest common divisor of 1000 and 1004), the main diagonal intersects one horizontal and one vertical line at the same time, which results in double-counting., so we must subtract 4 from the answer.
Therefore, the answer is 1000 + 1004 – 4 = 2000.

The solution is shown below.

Color the rhombuses based on their type and imagine the diagram represents a structure of small cubes arranged in a larger cube. If you look at the large cube from three different angles, you will see exactly the three types of rhombuses on the diagram.
Alternatively, the problem can be proven more rigorously by considering the three sets of non-intersecting broken lines connecting the pairs of opposite sides of the hexagon, as shown on the image below. The type of each rhombus is determined by the types of the broken lines passing through it. Therefore, there are n² rhombuses of each type, where n is the length of the hexagon’s sides.

Start from any point of the curve and keep moving along it, so that at each non-visited intersection you go over, until you get back to where you started from.

Connect the points in pairs of different colors so that the total length of all segments is minimal. If any two segments intersect, you can swap the two pairs among these four points and get a smaller total length.

The original puzzle has 12 unique solutions, up to rotation and symmetry. With the additional restriction imposed, there is only one solution.

This is the so called Simpson’s paradox. The reason it occurred is that during 1996 both players had high averages and Derek Jeter had many more hits than David Justice. In 1996 both players had low averages and David Justice had many more hits than Derek Jeter. You can see their official statistics below.
Player 1995 1996 1995-1996
Derek Jeter 12/48 (.250) 183/582 (.314)
195/630 (.310)
David Justice 104/411 (.253) 45/140 (.321) 149/551 (.270)

Each of the images in the first row depicts HAIR. Each of the images in the second row (except the second cell) depicts a BUG. Each of the images in the third row depicts ER.
The I from HAIR on the first row gets moved right after the B from BUG on the second row. Also, the U from BUG gets flipped upside down. Therefore, we get HARBINGER.