114*. What is the maximum number of natural numbers that can be written in a row so that the sum of any three consecutive numbers is even, and the sum of any four consecutive numbers is odd?
Let's denote the consecutive natural numbers of the row as $a_{1}, a_{2}, a_{3}$, and so on.

By the condition, the sums

$$
a_{1}+a_{2}+a_{3}, \quad a_{2}+a_{3}+a_{4}, \quad a_{3}+a_{4}+a_{5}, \quad a_{4}+a_{5}+a_{6}
$$

and others are even. By subtracting each sum, starting from the second, from the previous one, we get that the differences

$$
a_{4}-a_{1}, \quad a_{5}-a_{2}, \quad a_{6}-a_{3}, \ldots
$$

are even, and therefore, the pairs of numbers $a_{1}$ and $a_{4}, a_{2}$ and $a_{5}, a_{3}$ and $a_{6}$, etc., have the same parity.

Let's write down the odd sums consisting of four consecutive numbers:

$$
\begin{gathered}
a_{1}+a_{2}+a_{3}+a_{4}=\left(a_{1}+a_{2}+a_{3}\right)+a_{4} \\
a_{2}+a_{3}+a_{4}+a_{5}=\left(a_{2}+a_{3}+a_{4}\right)+a_{5} \\
a_{3}+a_{4}+a_{5}+a_{6}=\left(a_{3}+a_{4}+a_{5}\right)+a_{6}, \ldots
\end{gathered}
$$

From this, it follows that the numbers $a_{4}, a_{5}, a_{6}$, etc., are odd. But then the sum $a_{4}+a_{5}+a_{6}$ is odd, which contradicts the condition.

This contradiction arises every time the number of numbers is not less than 6. Let's try to take 5 numbers.

Reasoning similarly, we establish that the numbers $a_{4}$ and $a_{5}$ are odd, and therefore, by the previous, the numbers $a_{1}$ and $a_{2}$ are also odd. Then, since the sum $a_{1}+a_{2}+a_{3}$ is even, the number $a_{3}$ is even.

Let's do one more check and verify that if we take five numbers

$$
a_{1}, a_{2}, a_{3}, a_{4}, a_{5}
$$

where the number $a_{3}$ is even, and the others are odd, then each of the sums

$$
a_{1}+a_{2}+a_{3}, \quad a_{2}+a_{3}+a_{4}, \quad a_{3}+a_{4}+a_{5}
$$

is even, and each of the sums

$$
a_{1}+a_{2}+a_{3}+a_{4}, \quad a_{2}+a_{3}+a_{4}+a_{5}
$$

is odd.

Answer: 5