1077*. Find all natural numbers $k$ and $n$ for which the number

$$
\frac{7 k+15 n-1}{3 k+4 n}
$$

is a natural number.
$\triangle$ For natural $k$ and $n$, the inequality holds:

$$
2<\frac{7 k+15 n-1}{3 k+4 n}<4
$$

(verify!). Therefore, this fraction can only be equal to $3$:

$$
\frac{7 k+15 n-1}{3 k+4 n}=3, \quad 7 k+15 n-1=9 k+12 n, \quad 3 n-2 k=1
$$

To solve the last equation, express $k$ from it:

$$
k=\frac{3 n-1}{2}
$$

From this, it is clear that the number $n$ has only one restriction - it must be odd: $n=2 a-1(a \in N)$. Then

$$
k=\frac{3(2 a-1)-1}{2}=\frac{6 a-4}{2}=3 a-2
$$

Answer: $k=3 a-2, n=2 a-1(a \in N)$.