1. Start the search among 2-digit numbers. A dropdown number (DDN) with 2 digits is a number [tex]ab[/tex] such that
[tex]\dfrac{a+b}2 = a \implies a + b = 2a \implies b = a[/tex]
or
[tex]\dfrac{a+b}2 = b \implies a+b = 2b \implies a=b[/tex]
but both of these solutions violate the requirement that the digits are distinct, so there are no 2-digit DDNs.
A 3-digit DDN [tex]abc[/tex] is such that
[tex]\dfrac{a+b+c}3 = a \implies a+b+c = 3a \implies b+c = 2a[/tex]
or [tex]a+c=2b[/tex] if the average is [tex]b[/tex], or [tex]a+b=2c[/tex] if the average is [tex]c[/tex]. The smallest possible value for [tex]a[/tex] is 1 since we require 3 digits. Then [tex]b+c=2[/tex], and we can pick [tex]b=0[/tex] and [tex]c=2[/tex] to get the smallest DDN, 102.
2. In a 4-digit DDN [tex]abcd[/tex], we have
[tex]\dfrac{a+b+c+d}4 = a \implies a + b + c + d = 4a \implies b+c+d=3a[/tex]
or [tex]a+c+d=3b[/tex] or [tex]a+b+d=3c[/tex] or [tex]a+b+c=3d[/tex].
We're free to fix [tex]a=1[/tex] and [tex]b=0[/tex] to try to get the smallest DDN. This leaves us with [tex]c+d=3[/tex] or [tex]c+d=-1[/tex] or [tex]1+d=3c[/tex] or [tex]c=3d[/tex].
The first two cases are impossible - the only choices for [tex]c,d[/tex] such that [tex]c+d=3[/tex] are 1 and 2, and the sum of two positive integers must be positive. The smallest possible value of [tex]c[/tex] is 2; this leaves us with [tex]1+d=6[/tex] or [tex]2=3d[/tex], but the latter case is impossible because 3 does not divide 2. So [tex]d=5[/tex], and the smallest 4-digit DDN is 1025.
To find the largest DDN, start with the largest possible values for [tex]a[/tex] and [tex]b[/tex]. Let [tex]a=9[/tex] and [tex]b=8[/tex]. Then [tex]c+d=19[/tex] or [tex]c+d=15[/tex] or [tex]17+d=3c[/tex] or [tex]17+c=3d[/tex]. At most, we can have [tex]c+d=13[/tex] with 7 and 6, so the first two cases are impossible. If we maximize [tex]c=7[/tex], then either [tex]17+d=21\implies d=4[/tex] or [tex]24=3d\implies d=8[/tex] (which we don't want). So the largest 3-digit DDN is 9874.
3. I don't have an analytical solution to this, but using brute force (program) the total count is 112.
4. It is possible; consider 1249 and 1250, with digital averages
[tex]\dfrac{1+2+4+9}4=4 \text{ and } \dfrac{1+2+5+0}4=2[/tex]
which happens to be the smallest pair. (Also found with brute force.)
