Answer: 236

Solution:

The median is an integer that is not in the list, so the list cannot have odd length (then the median would be a member). The unique mode $9$ appears at least twice. Two items $9, 9$ sum to $18,$ not $30,$ so try four items $a \lt b \lt 9$ together with $9, 9,$ where $a$ and $b$ are distinct (a repeat would tie the mode) and $a + b = 12.$ The median $\frac{b + 9}{2}$ must be an integer, so $b$ is odd, and $a = 12 - b \lt b$ forces $b \gt 6.$ Thus $b = 7$ and $a = 5:$ the list $5, 7, 9, 9$ has median $8,$ which indeed does not appear.

No longer list works. With exactly two $9$s, six items would need four distinct other values summing to $12,$ namely $\{1, 2, 3, 6\}$ or $\{1, 2, 4, 5\},$ but both give median $4.5;$ eight or more items are impossible because six distinct positive values already sum to at least $21 \gt 12.$ With three $9$s the remaining items sum to $3,$ and every option either puts $9$ at the median or ties the mode. Four $9$s already sum to more than $30.$

The sum of squares is $25 + 49 + 81 + 81 = 236.$