The Fibonacci numbers are defined by $F_1=F_2=1$ and $F_n=F_{n-1}+F_{n-2}$ for $n>2$.  It is well-known that the sum of any $10$ consecutive Fibonacci numbers is divisible by $11$.  Determine the smallest integer $N$ so that the sum of any $N$ consecutive Fibonacci numbers is divisible by $12$.
1. **Define the Fibonacci sequence modulo 12:**
   The Fibonacci sequence is defined as:
   \[
   F_1 = 1, \quad F_2 = 1, \quad F_n = F_{n-1} + F_{n-2} \quad \text{for} \quad n > 2
   \]
   We need to consider this sequence modulo 12.

2. **Compute the first few terms of the Fibonacci sequence modulo 12:**
   \[
   \begin{aligned}
   F_1 &\equiv 1 \pmod{12}, \\
   F_2 &\equiv 1 \pmod{12}, \\
   F_3 &\equiv 2 \pmod{12}, \\
   F_4 &\equiv 3 \pmod{12}, \\
   F_5 &\equiv 5 \pmod{12}, \\
   F_6 &\equiv 8 \pmod{12}, \\
   F_7 &\equiv 1 \pmod{12}, \\
   F_8 &\equiv 9 \pmod{12}, \\
   F_9 &\equiv 10 \pmod{12}, \\
   F_{10} &\equiv 7 \pmod{12}, \\
   F_{11} &\equiv 5 \pmod{12}, \\
   F_{12} &\equiv 0 \pmod{12}, \\
   F_{13} &\equiv 5 \pmod{12}, \\
   F_{14} &\equiv 5 \pmod{12}, \\
   F_{15} &\equiv 10 \pmod{12}, \\
   F_{16} &\equiv 3 \pmod{12}, \\
   F_{17} &\equiv 1 \pmod{12}, \\
   F_{18} &\equiv 4 \pmod{12}, \\
   F_{19} &\equiv 5 \pmod{12}, \\
   F_{20} &\equiv 9 \pmod{12}, \\
   F_{21} &\equiv 2 \pmod{12}, \\
   F_{22} &\equiv 11 \pmod{12}, \\
   F_{23} &\equiv 1 \pmod{12}, \\
   F_{24} &\equiv 0 \pmod{12}.
   \end{aligned}
   \]

3. **Identify the period of the sequence modulo 12:**
   From the computed terms, we observe that the sequence repeats every 24 terms. This is the period of the Fibonacci sequence modulo 12.

4. **Sum of the first 24 terms modulo 12:**
   \[
   \sum_{k=1}^{24} F_k \equiv 1 + 1 + 2 + 3 + 5 + 8 + 1 + 9 + 10 + 7 + 5 + 0 + 5 + 5 + 10 + 3 + 1 + 4 + 5 + 9 + 2 + 11 + 1 + 0 \pmod{12}
   \]
   Simplifying the sum:
   \[
   \sum_{k=1}^{24} F_k \equiv 100 \pmod{12} \equiv 4 \pmod{12}
   \]
   This shows that the sum of the first 24 terms is not divisible by 12.

5. **Check sums of other lengths:**
   We need to find the smallest \( N \) such that the sum of any \( N \) consecutive Fibonacci numbers is divisible by 12. We test the sums of different lengths:
   \[
   \begin{aligned}
   \sum_{k=1}^{5} F_k &\equiv 1 + 1 + 2 + 3 + 5 \equiv 12 \equiv 0 \pmod{12}, \\
   \sum_{k=1}^{15} F_k &\equiv 1 + 1 + 2 + 3 + 5 + 8 + 1 + 9 + 10 + 7 + 5 + 0 + 5 + 5 + 10 \equiv 72 \equiv 0 \pmod{12}, \\
   \sum_{k=1}^{21} F_k &\equiv 1 + 1 + 2 + 3 + 5 + 8 + 1 + 9 + 10 + 7 + 5 + 0 + 5 + 5 + 10 + 3 + 1 + 4 + 5 + 9 + 2 \equiv 100 \equiv 4 \pmod{12}, \\
   \sum_{k=1}^{23} F_k &\equiv 1 + 1 + 2 + 3 + 5 + 8 + 1 + 9 + 10 + 7 + 5 + 0 + 5 + 5 + 10 + 3 + 1 + 4 + 5 + 9 + 2 + 11 + 1 \equiv 111 \equiv 3 \pmod{12}, \\
   \sum_{k=1}^{24} F_k &\equiv 1 + 1 + 2 + 3 + 5 + 8 + 1 + 9 + 10 + 7 + 5 + 0 + 5 + 5 + 10 + 3 + 1 + 4 + 5 + 9 + 2 + 11 + 1 + 0 \equiv 112 \equiv 4 \pmod{12}.
   \end{aligned}
   \]

6. **Conclusion:**
   The smallest \( N \) such that the sum of any \( N \) consecutive Fibonacci numbers is divisible by 12 is \( \boxed{24} \).