The figure below consists of several unit squares, $M$ of which are white and $N$ of which are green.  Compute $100M+N$.
[asy]
size(4cm);
int N = 4;
path square;
for (int x=-N; x<=N; ++x) {
  for (int y=-N+abs(x); y<=N-abs(x); ++y) {
    square = rotate(9)*((x+0.5,y+0.5)--(x+0.5,y-0.5)--(x-0.5,y-0.5)--(x-0.5,y+0.5)--cycle);
    if ((x+y) % 2 == 0) { filldraw(square, green, black); }
    else { filldraw(square, white, black); }
  }
}
[/asy]

[i]Proposed by Evan Chen[/i]
1. First, we need to understand the structure of the figure. The figure is composed of several unit squares arranged in a diamond shape. The diamond shape is symmetric about both the horizontal and vertical axes.

2. The diamond shape has a side length of 4 units, which means it extends 4 units from the center to each side. This results in a total of \(2 \times 4 + 1 = 9\) rows and columns of unit squares.

3. We need to count the number of white and green squares. The coloring pattern alternates between white and green squares, starting with a green square at the center.

4. To count the number of white and green squares, we can observe the pattern:
   - The center square is green.
   - The next layer around the center has 4 green squares and 4 white squares.
   - The next layer has 8 green squares and 8 white squares.
   - The outermost layer has 9 green squares and 9 white squares.

5. Summarizing the counts:
   - Center: 1 green
   - First layer: 4 green, 4 white
   - Second layer: 8 green, 8 white
   - Third layer: 9 green, 9 white

6. Adding up the counts:
   - Total green squares: \(1 + 4 + 8 + 9 = 22\)
   - Total white squares: \(4 + 8 + 9 = 21\)

7. However, the problem states that \(M\) is the number of white squares and \(N\) is the number of green squares. We need to verify the counts again.

8. Rechecking the counts:
   - The diamond shape has a total of \(9 \times 9 = 81\) squares.
   - Since the pattern alternates, half of the squares are green and half are white, with one extra green square at the center.

9. Correcting the counts:
   - Total green squares: \( \frac{81 + 1}{2} = 41\)
   - Total white squares: \( \frac{81 - 1}{2} = 40\)

10. Therefore, \(M = 40\) and \(N = 41\).

11. Finally, we compute \(100M + N\):
   \[
   100M + N = 100 \times 40 + 41 = 4000 + 41 = 4041
   \]

The final answer is \(\boxed{4041}\).