5. A list of numbers is created using the following rules:
- The first number is 3 and the second number is 4 .
- Each number after the second is the result of adding 1 to the previous number and then dividing by the number before that. In other words, for any three consecutive numbers in the list, $a, b, c$, we have $c=\frac{b+1}{a}$.
The rules tell us that the third number in the list is $\frac{4+1}{3}$ which equals $\frac{5}{3}$.
What is the smallest positive integer $N$ for which the sum of the first $N$ numbers in the list is equal to an odd integer that is greater than $2021 ?$
5. Using the given rules, the first 7 numbers in the list are
$$
\mathbf{3}, \mathbf{4}, \frac{5}{3}, \frac{(5 / 3)+1}{4}=\frac{8 / 3}{4}=\frac{2}{3}, \frac{(2 / 3)+1}{5 / 3}=\frac{5 / 3}{5 / 3}=\mathbf{1}, \frac{1+1}{2 / 3}=\frac{2}{2 / 3}=\mathbf{3}, \frac{3+1}{1}=\mathbf{4}
$$

Since the 6 th number equals the 1st number, the 7 th number equals the 2 nd number, and each number in the list depends on the previous two numbers, then the 8 th number equals the 3rd number, which will mean that the 9 th number will equal the 4 th number, and so on.
In other words, the list will repeat every 5 numbers since each group of 5 numbers is produced in exactly the same way as the previous 5 numbers.
The sum of the first 5 numbers is $3+4+\frac{5}{3}+\frac{2}{3}+1=8+\frac{7}{3}=10 \frac{1}{3}$.
Since $2021 \div 10 \frac{1}{3} \approx 195.6$, then 195 groups of 5 numbers will still have a sum less than 2021 .
In fact, since $195 \times 10 \frac{1}{3}=1950+65=2015$ and $195 \times 5=975$, then the sum of the first 975 numbers in the list is 2015 .
We make a chart of the next terms and the sum to that point until we reach a sum that is greater than 2021 and is an odd integer:
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline Term \# & 976 & 977 & 978 & 979 & 980 & 981 & 982 \\
\hline Term & 3 & 4 & $\frac{5}{3}$ & $\frac{2}{3}$ & 1 & 3 & 4 \\
\hline Sum to this point & 2018 & 2022 & $2023 \frac{2}{3}$ & $2024 \frac{1}{3}$ & $2025 \frac{1}{3}$ & $2028 \frac{1}{3}$ & $2032 \frac{1}{3}$ \\
\hline
\end{tabular}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline Term \# & 983 & 984 & 985 & 986 & 987 & 988 & 989 \\
\hline Term & $\frac{5}{3}$ & $\frac{2}{3}$ & 1 & 3 & 4 & $\frac{5}{3}$ & $\frac{2}{3}$ \\
\hline Sum to this point & 2034 & $2034 \frac{2}{3}$ & $2035 \frac{2}{3}$ & $2038 \frac{2}{3}$ & $2042 \frac{2}{3}$ & $2044 \frac{1}{3}$ & 2045 \\
\hline
\end{tabular}

Therefore, the smallest positive integer $N$ for which the sum of the first $N$ numbers is equal to an odd integer greater than 2021 is $N=989$ when the sum is 2045 .
ANSWER: 989