1. We need to count natural numbers $n$ such that $10 \le n \le 100$ and $3^n-3$ is divisible by $7$.

2. The divisibility condition is $3^n-3 \equiv 0 \pmod 7.$ So, $3^n \equiv 3 \pmod 7.$

3. Since $3$ is invertible modulo $7$, divide both sides by $3$: $3^{n-1} \equiv 1 \pmod 7.$

4. Now find the order of $3$ modulo $7$: [ 3^1 \equiv 3 \pmod 7, \quad 3^2=9 \equiv 2 \pmod 7, \quad 3^3=27 \equiv 6 \pmod 7, \quad 3^4=81 \equiv 4 \pmod 7, \quad 3^5=243 \equiv 5 \pmod 7, \quad 3^6=729 \equiv 1 \pmod 7. ] Hence the order of $3$ modulo $7$ is $6$.

5. Therefore, $3^{n-1} \equiv 1 \pmod 7$ iff $n-1 \equiv 0 \pmod 6.$ So, $n \equiv 1 \pmod 6.$

6. Now count numbers between $10$ and $100$ of the form $n=6k+1.$ The first such number $\ge 10$ is $13$, and the last such number $\le 100$ is $97$.

   So the sequence is $13,19,25,31,37,43,49,55,61,67,73,79,85,91,97.$

7. Number of terms: $\frac{97-13}{6}+1=\frac{84}{6}+1=14+1=15.$

Therefore, the required number of elements is $15.$