63 of 100: 40 Matchsticks

Consider first the outer $4 \times 4$ grid: ${\color{#D61F06}{\text{one}}}$ matchstick is required to be removed at a minimum, effectively eliminating one $1 \times 1$ square.

That leaves fifteen $1 \times 1$ squares. At most, how many squares can be eliminated by the removal of any given matchstick in the interior of the square grid? Two, since they share a common side. Thus, this will require $\left \lceil \dfrac {15}{2} \right \rceil = \color{#D61F06} \text{eight}$ matchsticks to be removed for a minimized total of ${\color{#D61F06}{1 + 8}} = \boxed{9}$ matchsticks removed.

All that is left is to show that this can be achieved:

Its nine.

Remove eight matchsticks to get rid of all inner squares.

Remove one more matchstick to get rid of the outer square.

Consider all the shaded squares . They don't have any mutual sides so we have to remove 1 matchstick per shaded square $\rightarrow$ 8 matchsticks should be removed to eliminate these 8 squares .

We also need to eliminate the other squares . The best way is that each removed matchstick is also a side of a white square $\rightarrow$ a removed matchstick must be a mutual side of a pair of (shaded , white) squares $\rightarrow$ put squares into pairs . That's why we see lots of pairs in all the solutions . To eliminate the biggest square (black sides) , we have to remove 1 more matchstick . Hence , the answer is 8 + 1 = $\boxed{9}$ .

By the way , while arranging 8 pairs , it seems impossible to eliminate 2 $\times$ 2 squares . In this case , replace 2 pairs by a 1 $\times$ 1 square and the yellow thing (I don't know what to call it) or a 3 $\times$ 1 rectangle . This 1 $\times$ 1 square is eliminated together with the biggest square , which means at least one of its sides is black .

edit : My old answer is wrong and this is my new answer :

Split the construction into symmetrical parts and you get 4 squares which can be further divided into 4 small square with a side of one match. To remove the square shapes we have to remove 2 matchsticks from each large square as such:

So that is 8 matchsticks so far. After that we also need to take into consideration the large square with a side of four matches which only requires the removal of 1 match.

8+1=9

Remove the second and fourth rows of matches to create two rows of rectangles. Then take one more match out of the perimeter to get rid of the big square.

We can see this problem as a n-2 x n-2 grid with a level of boxes around it such that an n x n grid is formed (A grid inside another grid). Suppose now we know how to solve the problem in the inner grid, say C {n-2} sticks to be removed. Now for the outer line we have 4n-4 boxes. Just divide them by pairs until one is left. We have then 2n-3 sticks in our hands. For the last pair note we have the outer box so we cut one of the little ones and the big, leaving one little box of the outer line intact. We have removed then 2n-2. This same method can be applied to the interior grid making the last cut in the method such that takes account of the box we left before. Then we have C {n} =2n-2 + C {n-2} . The base cases are clearly C {1} =1 and C {2} =3. This formula leads to 9 for n=4. This has not the structure of a rigurous proof but it does well as a conjecture. I'll be glad if you guys can spot any mistakes or build further.

Before we remove any matches, there are sixteen (1x1) squares. For each one, we need to remove one of its four sides. If we remove one of the inside matches, then we can remove two of the sixteen squares. So we can get all of the (1x1) squares by removing eight matches. However, we also need to remove one of the outside matches to get rid of the (4x4) square. Therefore, we need at least nine matches.