$\mathbf{N} 4$
For each positive integer $k$, let $t(k)$ be the largest odd divisor of $k$. Determine all positive integers $a$ for which there exists a positive integer $n$ such that all the differences
$$
t(n+a)-t(n), \quad t(n+a+1)-t(n+1), \quad \ldots, \quad t(n+2 a-1)-t(n+a-1)
$$
are divisible by 4 .
Answer. $\quad a=1,3$, or 5 .

Solution. A pair $(a, n)$ satisfying the condition of the problem will be called a winning pair. It is straightforward to check that the pairs $(1,1),(3,1)$, and $(5,4)$ are winning pairs.

Now suppose that $a$ is a positive integer not equal to 1,3 , and 5 . We will show that there are no winning pairs $(a, n)$ by distinguishing three cases.

Case 1: $a$ is even. In this case we have $a=2^{\alpha} d$ for some positive integer $\alpha$ and some odd $d$. Since $a \geq 2^{\alpha}$, for each positive integer $n$ there exists an $i \in\{0,1, \ldots, a-1\}$ such that $n+i=2^{\alpha-1} e$, where $e$ is some odd integer. Then we have $t(n+i)=t\left(2^{\alpha-1} e\right)=e$ and
$$
t(n+a+i)=t\left(2^{\alpha} d+2^{\alpha-1} e\right)=2 d+e \equiv e+2 \quad(\bmod 4) .
$$

So we get $t(n+i)-t(n+a+i) \equiv 2(\bmod 4)$, and $(a, n)$ is not a winning pair.
Case 2: $a$ is odd and $a>8$. For each positive integer $n$, there exists an $i \in\{0,1, \ldots, a-5\}$ such that $n+i=2 d$ for some odd $d$. We get
$$
t(n+i)=d \not \equiv d+2=t(n+i+4) \quad(\bmod 4)
$$
and
$$
t(n+a+i)=n+a+i \equiv n+a+i+4=t(n+a+i+4) \quad(\bmod 4) .
$$

Therefore, the integers $t(n+a+i)-t(n+i)$ and $t(n+a+i+4)-t(n+i+4)$ cannot be both divisible by 4 , and therefore there are no winning pairs in this case.
Case 3: $a=7$. For each positive integer $n$, there exists an $i \in\{0,1, \ldots, 6\}$ such that $n+i$ is either of the form $8 k+3$ or of the form $8 k+6$, where $k$ is a nonnegative integer. But we have
$$
t(8 k+3) \equiv 3 \not \equiv 1 \equiv 4 k+5=t(8 k+3+7) \quad(\bmod 4)
$$
and
$$
t(8 k+6)=4 k+3 \equiv 3 \not \equiv 1 \equiv t(8 k+6+7) \quad(\bmod 4) .
$$

Hence, there are no winning pairs of the form $(7, n)$.