, 2023

The answer is 2137.

First of all, let's observe that $x = 63$, $y = 2137$ satisfies the given relation. In fact,
$$\frac{S(63) + S(1)}{S(64)} = \frac{77 + 1}{100} = \frac{78}{100} < \frac{8}{10} \quad \text{and} \\ \frac{S(2137) + S(2023)}{S(4160)} = \frac{4131 + 3747}{10100} = \frac{7878}{10100} = \frac{78}{100} < \frac{8}{10}.$$

Now let's assume that the octal representation of $x$ ends with exactly $k$ digits “7” (maybe $k = 0$). In other words, $x = a \cdot 8^k + 8^k - 1$ where $a$ is a nonnegative integer satisfying $a \bmod 8 < 7$. In particular, we know that $S(a + 1) = S(a) + 1$ because there is no carry in addition in the octal system, since $a \bmod 8 < 7$. Therefore,
$$S(x) = S(a \cdot 8^k + 8^k - 1) = S(a) \cdot 10^k + \frac{7}{9}(10^k - 1) \quad \text{and} \\ S(x + 1) = S((a + 1) \cdot 8^k) = S(a + 1) \cdot 10^k = (S(a) + 1) \cdot 10^k.$$

Let's observe that $k = 0$ would imply that $S(x) + S(1) = S(x + 1)$ and therefore $k \neq 0$. $k \ge 3$ would imply $x \ge 8^3 - 1 > 100$ and therefore $k = 1$ or $k = 2$. If $k = 2$ then $a = 0$ (because $8^2 + 8^2 - 1 > 100$) and therefore $x = 8^2 - 1 = 63$ and $\frac{S(x)+S(1)}{S(x+1)} = \frac{78}{100}$. $k = 1$ is also impossible because:
$$\frac{10S(a) + 8}{10S(a) + 10} < \frac{8}{10} \implies 10S(a) + 8 < 8S(a) + 8 \implies S(a) < 0.$$

Now let's assume that we can find a solution with $y < 2137 = 4131_8$. Then $y$ has at most 4 digits in its octal representation. Let $d_i$ for $i = 0, 1, 2, 3$ be equal to 1 if there is a carry at the $i$-th position in addition of $y$ and $2023 = 3747_8$ in octal system. It is easy to conclude that
$$S(y+2023)-S(y)-S(2023) = \sum_{i=0}^{3} (-8) \cdot 10^i \cdot d_i + \sum_{i=0}^{3} 10^{i+1} \cdot d_i = 2 \sum_{i=0}^{3} 10^i \cdot d_i.$$
We have the implication:
$$\begin{aligned} \frac{S(y) + S(2023)}{S(y + 2023)} &= \frac{78}{100} & \implies & \frac{S(y) + S(2023)}{S(y + 2023) - S(y) - S(2023)} &= \frac{78}{22} \\ & \implies & S(y) + S(2023) &= \frac{78}{11} \sum_{i=0}^{3} 10^i \cdot d_i. \end{aligned}$$
We know that $d_3 = 1$ because otherwise we would have:
$$S(y) + S(2023) \le \frac{78}{11} \cdot 111 < 3747 = S(2023).$$
Because 78 is coprime with 11, we need to have the divisibility $11|\sum_{i=0}^{3} 10^i \cdot d_i$. Since $d_3 = 1$, there are only three possible cases: $\sum_{i=0}^{3} 10^i \cdot d_i = 1001, 1100, 1111$. They correspond to $S(y) + S(2023) = 7098, 7800, 7878$, so $S(y) = 3351, 4053, 4131$. The first two cases are impossible because they assume $d_1 = 0$ but $5 + 4 > 7$, so there has to be a carry at the first position.