1. Let the three-digit number be written as $abc$, where:

   • $a$ is the hundreds digit,
   • $b$ is the tens digit,
   • $c$ is the units digit.

   We need: $a+b+c=15$

2. Also, the number must lie between $212$ and $999$.

   Since every number from $212$ to $999$ is a three-digit number, we count valid digit triples $(a,b,c)$ with: $1\le a\le 9,\quad 0\le b,c\le 9$ and then exclude those less than $212$.

3. First count all three-digit numbers whose digits sum to $15$.

   We solve: $a+b+c=15$ with $a\ge 1$, $b,c\ge 0$.

   Put: $a'=a-1\ge 0$ Then: $a'+b+c=14$

   Number of non-negative integer solutions is: $\binom{14+3-1}{3-1}=\binom{16}{2}=120$

4. Now check digit upper bounds $a,b,c\le 9$.

   Since $a+b+c=15$, at most one variable can exceed $9$.

   • If $a\ge 10$, let $a''=a-10\ge 0$. Then: $a''+b+c=5$ Number of solutions: $\binom{5+3-1}{2}=\binom{7}{2}=21$

   • If $b\ge 10$, let $b''=b-10\ge 0$. Then: $a+b''+c=5$ with $a\ge 1$. Put $a'=a-1\ge 0$: $a'+b''+c=4$ Number of solutions: $\binom{4+3-1}{2}=\binom{6}{2}=15$

   • If $c\ge 10$, similarly number of solutions is: $15$

   Hence valid three-digit numbers with digit sum $15$ are: $120-21-15-15=69$

5. Now exclude those numbers less than $212$.

   Such numbers are from $100$ to $211$. We need digit sum $15$.

   • If $a=1$, then: $b+c=14$ Possible digit pairs are: $(5,9),(6,8),(7,7),(8,6),(9,5)$ giving numbers: $159,168,177,186,195$ These are all $<212$.

   • If $a=2$, then for numbers $<212$, we only have $200$ to $211$. Their digit sums are at most $2+1+1=4$, so none has digit sum $15$.

   Therefore, exactly $5$ such numbers are below $212$.

6. Required count: $69-5=64$

Therefore, the number of natural numbers between $212$ and $999$ whose digits sum to $15$ is: $\boxed{64}$