$77$ stones weighing $1,2,\dots, 77$ grams are divided into $k$ groups such that total weights of each group are different from each other and each group contains less stones than groups with smaller total weights. For how many $k\in \{9,10,11,12\}$, is such a division possible?

$ 
\textbf{(A)}\ 4
\qquad\textbf{(B)}\ 3
\qquad\textbf{(C)}\ 2
\qquad\textbf{(D)}\ 1
\qquad\textbf{(E)}\ \text{None of above}
$
1. **Calculate the total sum of the weights of the stones:**
   \[
   T = 1 + 2 + 3 + \cdots + 77 = \frac{77 \cdot (77 + 1)}{2} = \frac{77 \cdot 78}{2} = 3003
   \]

2. **Determine the weight of the heaviest group:**
   \[
   W_k = \left\lceil \frac{T}{k} \right\rceil = \left\lceil \frac{3003}{k} \right\rceil
   \]

3. **Determine the number of stones in the heaviest group:**
   \[
   N_k = \left\lceil \frac{W_k}{77} \right\rceil
   \]

4. **Ensure that the total number of stones in all groups is at most 77:**
   \[
   N_k + (N_k + 1) + (N_k + 2) + \cdots + (N_k + k - 1) = k \cdot \left(N_k + \frac{k-1}{2}\right) \leq 77
   \]

5. **Analyze each value of \( k \in \{9, 10, 11, 12\} \):**

   - For \( k = 12 \):
     \[
     W_{12} = \left\lceil \frac{3003}{12} \right\rceil = \left\lceil 250.25 \right\rceil = 251
     \]
     \[
     N_{12} = \left\lceil \frac{251}{77} \right\rceil = \left\lceil 3.26 \right\rceil = 4
     \]
     \[
     12 \cdot \left(4 + \frac{12-1}{2}\right) = 12 \cdot \left(4 + 5.5\right) = 12 \cdot 9.5 = 114 > 77
     \]
     Therefore, \( k = 12 \) does not satisfy the condition.

   - For \( k = 11 \):
     \[
     W_{11} = \left\lceil \frac{3003}{11} \right\rceil = \left\lceil 273 \right\rceil = 273
     \]
     \[
     N_{11} = \left\lceil \frac{273}{77} \right\rceil = \left\lceil 3.54 \right\rceil = 4
     \]
     \[
     11 \cdot \left(4 + \frac{11-1}{2}\right) = 11 \cdot \left(4 + 5\right) = 11 \cdot 9 = 99 > 77
     \]
     Therefore, \( k = 11 \) does not satisfy the condition.

   - For \( k = 10 \):
     \[
     W_{10} = \left\lceil \frac{3003}{10} \right\rceil = \left\lceil 300.3 \right\rceil = 301
     \]
     \[
     N_{10} = \left\lceil \frac{301}{77} \right\rceil = \left\lceil 3.91 \right\rceil = 4
     \]
     \[
     10 \cdot \left(4 + \frac{10-1}{2}\right) = 10 \cdot \left(4 + 4.5\right) = 10 \cdot 8.5 = 85 > 77
     \]
     Therefore, \( k = 10 \) does not satisfy the condition.

   - For \( k = 9 \):
     \[
     W_{9} = \left\lceil \frac{3003}{9} \right\rceil = \left\lceil 333.67 \right\rceil = 334
     \]
     \[
     N_{9} = \left\lceil \frac{334}{77} \right\rceil = \left\lceil 4.34 \right\rceil = 5
     \]
     \[
     9 \cdot \left(5 + \frac{9-1}{2}\right) = 9 \cdot \left(5 + 4\right) = 9 \cdot 9 = 81 > 77
     \]
     Therefore, \( k = 9 \) does not satisfy the condition.

Since none of the values \( k \in \{9, 10, 11, 12\} \) satisfy the condition, the answer is:

The final answer is $\boxed{\text{(E)}}$