N16 (17-4, USSR) Let $A$ be the sum of the digits of the decimal number $4444^{4144}$, and let $B$ be the sum of the digits of $A$. Find the sum of the digits of $B$ (all numbers here are in decimal).

Let's translate the problem and the solution step by step.

### Problem:
N16 (17-4, USSR) Let \( A \) be the sum of the digits of the decimal number \( 4444^{4144} \), and let \( B \) be the sum of the digits of \( A \). Find the sum of the digits of \( B \) (all numbers here are in decimal).

### Solution:
1. **Estimate the number of digits in \( 4444^{4144} \):**
   - The number of digits \( d \) of a number \( n \) can be approximated by \( d = \lfloor \log_{10} n \rfloor + 1 \).
   - For \( 4444^{4144} \):
     \[
     \log_{10} (4444^{4144}) = 4144 \log_{10} 4444
     \]
   - Using \( \log_{10} 4444 \approx 3.6478 \):
     \[
     4144 \log_{10} 4444 \approx 4144 \times 3.6478 \approx 15120.5552
     \]
   - Therefore, the number of digits is:
     \[
     d = \lfloor 15120.5552 \rfloor + 1 = 15121
     \]

2. **Estimate the sum of the digits \( A \):**
   - The sum of the digits of a number with \( d \) digits, where each digit is at most 9, is at most \( 9d \).
   - For \( 4444^{4144} \):
     \[
     A \leq 9 \times 15121 = 136089
     \]

3. **Estimate the sum of the digits \( B \):**
   - The number \( A \) is at most 136089, which has 6 digits.
   - The sum of the digits of a 6-digit number, where each digit is at most 9, is at most \( 9 \times 6 = 54 \).
   - Therefore, \( B \leq 54 \).

4. **Find the sum of the digits of \( B \):**
   - The number \( B \) is at most 54, which has 2 digits.
   - The sum of the digits of a 2-digit number, where each digit is at most 9, is at most \( 9 + 9 = 18 \).
   - Therefore, the sum of the digits of \( B \) is at most 18.

5. **Use modular arithmetic to find the exact sum:**
   - The sum of the digits of a number is congruent to the number itself modulo 9.
   - Calculate \( 4444^{4144} \mod 9 \):
     \[
     4444 \equiv 7 \pmod{9}
     \]
     \[
     4444^{4144} \equiv 7^{4144} \pmod{9}
     \]
   - Since \( 7^3 \equiv 1 \pmod{9} \):
     \[
     7^{4144} = 7^{3 \cdot 1381 + 1} = (7^3)^{1381} \cdot 7 \equiv 1^{1381} \cdot 7 \equiv 7 \pmod{9}
     \]
   - Therefore, \( 4444^{4144} \equiv 7 \pmod{9} \).

6. **Sum of the digits of \( A \) and \( B \):**
   - Since \( A \equiv 7 \pmod{9} \) and \( B \equiv 7 \pmod{9} \), the sum of the digits of \( B \) must also be 7.

### Final Answer:
\[
\boxed{7}
\]
Solving: Since
$$
4444^{1444}<10000^{1414}=10^{4 \times 1444},
$$

the number of digits of $4444^{4444}$ does not exceed
$$
4 \times 4444=17776.
$$
$$
\begin{array}{l} 
\therefore \quad A \leqslant 17776 \times 9=159984. \\
B \leqslant 1+5 \times 9=46,
\end{array}
$$

so the sum of the digits of $B$, $C \leqslant 4+9=13$.
Since a number is congruent to its sum of digits modulo 9, we have
$$
\begin{aligned}
C & \equiv B \equiv A \equiv 4444^{4444} \\
& \equiv 7^{4444}=\left(7^{3}\right)^{1481+1} \\
& \equiv 1^{1481} \times 7=7(\bmod 9),
\end{aligned}
$$

Given $C \leqslant 13$ and $C \equiv 7(\bmod 9)$, we know $C=7$.
Therefore, the sum of the digits of $B$ is 7.