Let $N = 123456789101112\dots4344$ be the $79$-digit number obtained that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$?

$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 44$
To find the remainder of \( N \) when it is divided by \( 45 \), we need to consider the remainders when \( N \) is divided by \( 5 \) and \( 9 \) separately, and then use the Chinese Remainder Theorem (CRT) to combine these results.

1. **Finding \( N \mod 5 \):**
   - The last digit of \( N \) is \( 4 \) (since \( N \) ends with \( 44 \)).
   - Therefore, \( N \equiv 4 \pmod{5} \).

2. **Finding \( N \mod 9 \):**
   - The remainder when \( N \) is divided by \( 9 \) can be found by summing the digits of \( N \) and then taking that sum modulo \( 9 \).
   - We need to sum the digits of all numbers from \( 1 \) to \( 44 \).

   Let's break this down:
   - Sum of digits from \( 1 \) to \( 9 \):
     \[
     1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
     \]
   - Sum of digits from \( 10 \) to \( 19 \):
     \[
     1 + 0 + 1 + 1 + 1 + 2 + 1 + 3 + 1 + 4 + 1 + 5 + 1 + 6 + 1 + 7 + 1 + 8 + 1 + 9 = 46
     \]
   - Sum of digits from \( 20 \) to \( 29 \):
     \[
     2 + 0 + 2 + 1 + 2 + 2 + 2 + 3 + 2 + 4 + 2 + 5 + 2 + 6 + 2 + 7 + 2 + 8 + 2 + 9 = 47
     \]
   - Sum of digits from \( 30 \) to \( 39 \):
     \[
     3 + 0 + 3 + 1 + 3 + 2 + 3 + 3 + 3 + 4 + 3 + 5 + 3 + 6 + 3 + 7 + 3 + 8 + 3 + 9 = 48
     \]
   - Sum of digits from \( 40 \) to \( 44 \):
     \[
     4 + 0 + 4 + 1 + 4 + 2 + 4 + 3 + 4 + 4 = 30
     \]

   Adding these sums together:
   \[
   45 + 46 + 47 + 48 + 30 = 216
   \]

   - Now, we take \( 216 \mod 9 \):
     \[
     216 \div 9 = 24 \quad \text{(remainder 0)}
     \]
     Therefore, \( 216 \equiv 0 \pmod{9} \).

3. **Using the Chinese Remainder Theorem (CRT):**
   - We have:
     \[
     N \equiv 4 \pmod{5}
     \]
     \[
     N \equiv 0 \pmod{9}
     \]

   - We need to find \( x \) such that:
     \[
     x \equiv 4 \pmod{5}
     \]
     \[
     x \equiv 0 \pmod{9}
     \]

   - Since \( x \equiv 0 \pmod{9} \), we can write \( x = 9k \) for some integer \( k \).
   - Substituting into the first congruence:
     \[
     9k \equiv 4 \pmod{5}
     \]
     \[
     9 \equiv 4 \pmod{5} \quad \text{(since 9 mod 5 is 4)}
     \]
     \[
     4k \equiv 4 \pmod{5}
     \]
     \[
     k \equiv 1 \pmod{5}
     \]
     \[
     k = 5m + 1 \quad \text{for some integer } m
     \]

   - Therefore:
     \[
     x = 9k = 9(5m + 1) = 45m + 9
     \]
     \[
     x \equiv 9 \pmod{45}
     \]

Thus, the remainder when \( N \) is divided by \( 45 \) is \( 9 \).

The final answer is \(\boxed{9}\)