*Thank you to Neal for sending in his thoughts about Equal Averages*

I think the answer is 33.

Reasoning. Start with a,b,100,c,100+a in increasing order.

This covers median and range constraints

To get a mode of 100, either b=100 or c=100

So consider 2 cases:

1) Try b=100:

So the numbers are

a,100,100,c,100+a

But the mean is 100 also

So a+100+100+c+100+a=500

so c=200-2a

But also, 100<=c<=100+a

so 100<=200-2a<=100+a

this gives 34<=a<=50

i.e. a set of 17 groups

from 34,100,100,132,134

to 50,100,100,100,150

2) try b=100

So the numbers are

a,b,100,100,100+a

progressing similar to above

The mean is (a+b+100+100+100+a)/5 = 100

so b=200-2a

However, a<=b<=100

so a<=200-2a<=100

so 50<=a<=66

This gives a range of 17 answers, from 50,100,100,100,150 up to 66,68,100,100,166

Finally. combine the two cases

(a,100,100,200-2a,100+a) for 34<=a<=50

and

(a,200-2a,100,100,100+a) for 50<=a<66

Which is 33 groups (because one group (50,100,100,100,150) is shared by both cases