The three values above the median must be the three greatest values. Express all three using $n$.

Use the mean to find the required sum of the two values between the minimum and the median. Then compare that sum with the smallest possible sum those two distinct integers could have.

Algebra is the fastest method because the conditions require reasoning about ordered distinct integers. Desmos can verify the bound once the expressions are set up.

Enter $y=80-4x$ and $y=2x-29$. Here, $x$ represents the greatest value, the first expression is the required sum of the two unknown middle values, and the second is their smallest possible sum. The first graph must be at or above the second graph.

The range and positivity require $x\ge17$. The graphs intersect between $18$ and $19$, so inspect the whole-number candidates allowed by both conditions. To verify the largest candidate, enter $L=[2,3,5,9,16,17,18]$, then evaluate $\operatorname{mean}(L)$ and $\max(L)-\min(L)$.

Let $n$ be the greatest value. Because the range is $16$, the least value is $n-16$. The three values greater than the median are consecutive, so they are $n-2$, $n-1$, and $n$. Let the two remaining values be $a$ and $b$, so the data set has the form $n-16,a,b,9,n-2,n-1,n$.

A mean of $10$ for $7$ values means that the sum is $70$. Therefore, $(n-16)+a+b+9+(n-2)+(n-1)+n=70$, which simplifies to $a+b=80-4n$.

The least value, $n-16$, must be positive, so $n\ge17$. Also, since all values are distinct positive integers, the smallest possible values of $a$ and $b$ are $n-15$ and $n-14$. Thus, $a+b\ge2n-29$. Combining this with $a+b=80-4n$ gives $80-4n\ge2n-29$, so $n\le\frac{109}{6}$. Because $n$ is an integer, $n\le18$.

The data set $2,3,5,9,16,17,18$ has sum $70$, so its mean is $10$. Its median is $9$, its range is $18-2=16$, and its three values greater than $9$ are consecutive. Therefore, the greatest possible value is $18$.