778*. Find the smallest natural number that can be represented in exactly two ways as $3 x+4 y$, where $x$ and $y$ are natural numbers.
$\triangle$ Let

$$
3 x+4 y=n
$$

where $n-$ is a natural number. Let's solve this equation first in integers:

$$
\begin{array}{cl}
3 x=n-4 y, & x=\frac{n-4 y}{3}=-y+\frac{n-y}{3} \\
\frac{n-y}{3}=t \in Z ; \quad n-y=3 t, \quad y=n-3 t ; \quad x=-(n-3 t)+t=4 t-n
\end{array}
$$

Next, let's find all $t$ for which $x$ and $y$ are positive:

$$
\left\{\begin{array}{l}
4 t-n>0 \\
n-3 t>0
\end{array}\right.
$$

From this, $\frac{n}{4}<t<\frac{n}{3}$.

Now, let's consider that, first, $t$ must be an integer, and second, the interval $\left(\frac{n}{4}, \frac{n}{3}\right)$ must contain exactly two integer points.

Let's think: what can be the length of this interval?

If it is greater than 3, then, obviously, the interval contains at least three integer points. If the length of the interval does not exceed 3 but is greater than 2, then the interval contains two or three such points (examples can be the intervals $(2,4 ; 4,9)$ and $(2,9; 5,3)$). If the length of the interval does not exceed 2 but is greater than 1, then it contains two or one integer point (examples are the intervals $(4,7; 6,1)$ and $(4,1; 5,9)$). If the length does not exceed 1, then the interval contains one or even no integer points.

Therefore, the length of the interval of interest must be within the limits from 1 to 3, including 3 but excluding 1.

The length of the interval $\left(\frac{n}{4}, \frac{n}{3}\right)$ is $\frac{n}{3}-\frac{n}{4}=\frac{n}{12}$. Then

$$
1<\frac{n}{12} \leqslant 3, \quad 12<n \leqslant 36
$$

Now let's perform a check of the values of $n$, starting with $n=13$.

If $n=13$, then the inequality $\frac{n}{4}<t<\frac{n}{3}$ takes the form:

$$
\frac{13}{4}<t<\frac{13}{3}, \quad 3 \frac{1}{4}<t<4 \frac{1}{3}
$$

But this interval contains only one integer.

If $n=14$, then

$$
\frac{14}{4}<t<\frac{14}{3}, \quad 3 \frac{1}{2}<t<4 \frac{2}{3} .
$$

And this case should be discarded.

The same picture is obtained for $n=15,16,17,18$ (check it!).

And only for $n=19$ does it change:

$$
\frac{19}{4}<t<\frac{19}{3}, \quad 4 \frac{3}{4}<t<6 \frac{1}{3}
$$

and this interval contains two integer points - 5 and 6.

Answer: 19.