No.

Given sets
$$
a_1,a_2,\dots,a_{99},b_1{\rm\ and\ }a_1,a_2,\dots,a_{99},b_2
$$
we see that $b_1$ and $b_2$ must be the same color, say, red. Then from
$$
b_1,b_2,c_1,c_2,\dots,c_{98}{\rm\ and\ }d_1,d_2,c_1,c_2,\dots,c_{98}
$$
we see $d_1$ and $d_2$ must both be red. Then from
$$
d_1,d_2,e_1,e_2,\dots,e_{98}{\rm\ and\ }f_1,f_2,e_1,e_2,\dots,e_{98}
$$
we see that $f_1$ and $f_2$ must both be red. Dot, dot, dot. You wind up with as many elements as you like, all of which must be red, and none of them are in more than two of the sets. Once you have more than 50 of them, you can put them in another set which will then have more than 50 red points.

Obviously, we can take 100 to be a variable in this problem and solution, provided we restrict its range to the positive even integers and understand 50 to be 100/2.