How many 7-element subsets of $\{1, 2, 3,\ldots , 14\}$ are there, the sum of whose elements is divisible by $14$?
1. **Define the Polynomial:**
   Consider the polynomial 
   \[
   f(x) = (1 + x)(1 + x^2)(1 + x^3) \cdots (1 + x^{14}).
   \]
   Each term in the expansion of this polynomial corresponds to a unique subset of $\{1, 2, 3, \ldots, 14\}$, and the exponent of $x$ in each term corresponds to the sum of the elements in that subset.

2. **Roots of Unity Filter:**
   To find the number of 7-element subsets whose sum is divisible by 14, we use the roots of unity filter. Specifically, we need to evaluate the sum of the coefficients of $x^k$ in $f(x)$ where $k$ is a multiple of 14. This can be done by evaluating $f(x)$ at the 14th roots of unity.

3. **Evaluate at Roots of Unity:**
   Let $\zeta$ be a primitive 14th root of unity. The 14th roots of unity are $1, \zeta, \zeta^2, \ldots, \zeta^{13}$. We need to compute:
   \[
   \frac{1}{14} \sum_{k=0}^{13} f(\zeta^k).
   \]
   This sum will give us the number of subsets whose sum is divisible by 14.

4. **Simplify the Polynomial:**
   Notice that $f(x)$ can be written as:
   \[
   f(x) = \prod_{k=1}^{14} (1 + x^k).
   \]
   When evaluated at the 14th roots of unity, we have:
   \[
   f(\zeta^k) = \prod_{j=1}^{14} (1 + (\zeta^k)^j).
   \]
   Since $\zeta$ is a 14th root of unity, $(\zeta^k)^j = \zeta^{kj}$ cycles through the roots of unity.

5. **Sum of Evaluations:**
   The sum of the evaluations at the roots of unity is:
   \[
   \sum_{k=0}^{13} f(\zeta^k).
   \]
   By symmetry and properties of roots of unity, this sum simplifies to:
   \[
   \sum_{k=0}^{13} f(\zeta^k) = 14 \cdot \text{(number of 7-element subsets whose sum is divisible by 14)}.
   \]

6. **Final Calculation:**
   The number of 7-element subsets of $\{1, 2, 3, \ldots, 14\}$ is given by $\binom{14}{7}$. Therefore, the number of such subsets whose sum is divisible by 14 is:
   \[
   \frac{1}{14} \sum_{k=0}^{13} f(\zeta^k) = \frac{1}{14} \cdot \binom{14}{7}.
   \]

7. **Compute the Binomial Coefficient:**
   \[
   \binom{14}{7} = \frac{14!}{7! \cdot 7!} = 3432.
   \]

8. **Final Answer:**
   \[
   \frac{3432}{14} = 245.
   \]

The final answer is $\boxed{245}$.