Set the interval of colored flags to X.

Because only four flags have not moved,

So 800/4=200m, that is, every 200m, the flags inserted are the same, only one flag has not moved.

Therefore, the minimum common multiple of the interval between re-inserted signs and the original interval of 50 meters must be 200.

Because there are more flags now than before.

So x < 50

So x=40.

2、

2= 1×2

2+4=6=2×3

2+4+6= 12=3×4

2+4+6+8==20=4×5

2+4+6......+98+ 100=(2550 )=(50 )×(5 1 )

......

2+4+6+8+......+N=(( (n∧2+2n)/4)=(n/2 )×(n+2)/2