For which positive integer values of $k$ will the expression

$$
N=3^{6 n-1}-k \cdot 2^{3 n-2}+1
$$

be divisible by 7 for every positive integer value of $n$?
If $N$ is divisible by 7 for some value of $k$ for all $n$, then it is also divisible by 7 for $n=1$, so

$$
3^{5}-2 k+1=244-2 k=7 m
$$

Here, the left side is even, so the right side must also be even. This holds if and only if $m$ is even: $m=2 m_{1}$. Thus, for $k$ we get

$$
k=122-7 m_{1}=7\left(17-m_{1}\right)+3=7 m_{2}+3
$$

Here, $m_{2}$ can take any integer value, and (1) holds with $m=34-2 m_{2}$.

We will show from the other side that $N$ differs from the value corresponding to $n=1$ by a number divisible by 7 for all $n$. Indeed,

$$
N-\left(3^{5}-2 k+1\right)=3^{5}\left(3^{6(n-1)}-1\right)-2 k\left(2^{3(n-1)}-1\right)
$$

and the first difference on the right side is divisible by $3^{6}-1=\left(3^{3}-1\right)\left(3^{3}+1\right)=26 \cdot 28$, and thus also by 7, while the second term is divisible by $2^{3}-1=7$, so the entire difference is.

Thus, we have found that $N$ is divisible by 7 for all $n$ if and only if this holds for the value corresponding to $n=1$, and for this, it is necessary and sufficient that $k$ gives a remainder of 3 when divided by 7.