Let the three-digit integer be $\overline{abc}$, where:

• $a$ is the hundreds digit, so $1 \le a \le 9$
• $b,c$ are the tens and units digits, so $0 \le b,c \le 9$

We need: $a+b+c=14$

We must count the number of digit triples $(a,b,c)$ satisfying these conditions.

1. Remove the lower bound on $a$

Since $a\ge 1$, define $x=a-1$ Then $x\ge 0$ and $x\le 8$.

The equation becomes: $(x+1)+b+c=14$ $x+b+c=13$

Now we count non-negative integer solutions of $x+b+c=13$ with digit restrictions: $x\le 8,\quad b\le 9,\quad c\le 9$

2. Count all non-negative solutions without upper bounds

The number of non-negative integer solutions of $x+b+c=13$ is $\binom{13+3-1}{3-1}=\binom{15}{2}=105$

3. Subtract invalid cases

We now subtract solutions where any variable exceeds its allowed maximum.

Case 1: $x\ge 9$

Let $x'=x-9\ge 0$ Then $x'+b+c=4$ Number of solutions: $\binom{4+3-1}{2}=\binom{6}{2}=15$

Case 2: $b\ge 10$

Let $b'=b-10\ge 0$ Then $x+b'+c=3$ Number of solutions: $\binom{3+3-1}{2}=\binom{5}{2}=10$

Case 3: $c\ge 10$

Similarly, number of solutions: $10$

4. Check overlaps

• $x\ge 9$ and $b\ge 10$ would require at least $9+10=19 > 13$, impossible.
• $x\ge 9$ and $c\ge 10$ impossible.
• $b\ge 10$ and $c\ge 10$ impossible.

So there are no overlaps to add back.

5. Final count

Thus required number is $105-15-10-10=70$

Therefore, the number of three-digit integers whose digits sum to $14$ is $\boxed{70}$