We are given $X = \{n \in \mathbb{N} : 1 \le n \le 50\} = \{1,2,3,\dots,50\}.$

Define the sets:

• $A = \{n \in X : n \text{ is a multiple of } 2\}$
• $B = \{n \in X : n \text{ is a multiple of } 7\}$

The smallest subset of $X$ containing both $A$ and $B$ is their union: $A \cup B.$

So we need to find $|A \cup B|.$ Using the inclusion-exclusion principle: $|A \cup B| = |A| + |B| - |A \cap B|.$

Count multiples of $2$ from $1$ to $50$: $|A| = \left\lfloor \frac{50}{2} \right\rfloor = 25.$

Count multiples of $7$ from $1$ to $50$: $|B| = \left\lfloor \frac{50}{7} \right\rfloor = 7.$ These are $7,14,21,28,35,42,49$.

Count numbers that are multiples of both $2$ and $7$. Such numbers are multiples of $\operatorname{lcm}(2,7)=14.$ Hence, $|A \cap B| = \left\lfloor \frac{50}{14} \right\rfloor = 3.$ These are $14,28,42$.

Therefore, $|A \cup B| = 25 + 7 - 3 = 29.$

So the number of elements in the required smallest subset is $\boxed{29}.$

Comparing with the stored correct answer $29$, they match.