1. We need all six-digit numbers of the form $x_1x_2x_3x_4x_5x_6$ where $0<x_1<x_2<x_3<x_4<x_5<x_6.$

   Since the digits are strictly increasing and $x_1>0$, the digits must be chosen from $\{1,2,3,4,5,6,7,8,9\}.$

2. For any choice of 6 distinct digits from $\{1,2,\dots,9\}$, there is exactly one such number: write them in increasing order.

   Hence the numbers in increasing order correspond exactly to the 6-element subsets of $\{1,2,\dots,9\}$ in lexicographic order.

3. Let us find the $72^{\text{nd}}$ such number.

Step 1: Fix the first digit

If the first digit is $1$, then we choose the remaining 5 digits from $\{2,3,\dots,9\}$. Number of such choices: $\binom{8}{5}=56.$

So positions $1$ to $56$ begin with $1$.

Since $72>56$, the $72^{\text{nd}}$ number does not begin with $1$. It lies among numbers beginning with $2$.

Its position among numbers starting with $2$ is $72-56=16.$

Step 2: Fix the second digit

Now consider numbers starting with $2$.

If the second digit is $3$, then choose remaining 4 digits from $\{4,5,6,7,8,9\}$. Count: $\binom{6}{4}=15.$

Thus, among numbers starting with $2$, positions $1$ to $15$ are of the form $23\_\_\_\_.$

We need the $16^{\text{th}}$ among those starting with $2$, so it is not of the form $23\cdots$. Therefore the second digit must be $4$.

Now its position among numbers starting with $24$ is $16-15=1.$

Step 3: Smallest number starting with $24$

To get the first number beginning with $24$, choose the smallest possible remaining digits: $5,6,7,8.$

So the number is $245678.$

Thus the $72^{\text{nd}}$ number is $245678.$

4. Sum of its digits: $2+4+5+6+7+8=32.$

Therefore, the required sum is $\boxed{32}.$