1. Let the seven-digit number use digits only from $\{1,2,3,4\}$.

   If the digits are $x_1,x_2,\dots,x_7$, then $x_i \in \{1,2,3,4\}$ and $x_1+x_2+\cdots+x_7=12.$

2. Since each digit is at least $1$, subtract $1$ from each digit: $y_i=x_i-1.$ Then $y_i\in \{0,1,2,3\}$ and $y_1+y_2+\cdots+y_7 = 12-7 = 5.$

3. Now we need the number of nonnegative integer solutions of $y_1+y_2+\cdots+y_7=5,$ with the restriction $y_i\le 3$.

4. First count all nonnegative solutions without the upper bound: By stars and bars, $\binom{5+7-1}{7-1}=\binom{11}{6}=462.$

5. Subtract the invalid solutions where some $y_i\ge 4$.

   Let $y_i' = y_i-4 \ge 0$ for one such variable. Then the equation becomes $y_i' + \text{(other 6 variables)} = 1.$ Number of solutions for a fixed $i$: $\binom{1+7-1}{7-1}=\binom{7}{6}=7.$

   Since there are $7$ choices of $i$, total invalid solutions counted are $7\times 7 = 49.$

6. Can two variables both be $\ge 4$?

   No, because that would require sum at least $8$, but total sum is only $5$ for the $y_i$'s. So no overlap cases exist.

7. Therefore valid solutions: $462-49=413.$

8. Each valid solution corresponds uniquely to a seven-digit number, because each ordered 7-tuple $(x_1,\dots,x_7)$ gives exactly one number.

Hence, the required number of seven-digit integers is $\boxed{413}.$