$23x + 43y = 3000$

Rearranging $23x + 43y = 3000$ gives $y = \frac{3000 - 23x}{43}$ which can be further rearranged to $y = 70 - \frac{10}{43} - x + \frac{20x}{43}$ and again to $y = 70 - x + 10(\frac{2x - 1}{43})$ . For $y$ to be a natural number, $2x - 1$ must be divisible by $43$ .

Since $x$ is a natural number, $2x - 1$ must be odd, and so $\frac{2x - 1}{43}$ must be odd, and so $2x - 1 = 43(2n + 1)$ for any natural number $n$ , which simplifies to $x = 43n + 22$ .

When $y = 0$ , $23x + 43 \cdot 0 = 3000$ solves to $\frac{3000}{23} \approx 130.4$ . Since $x$ is a natural number it must be an integer between $0$ and $130$ , and the only values of $x = 43n + 22$ for any natural number $n$ in this range are $22$ , $65$ , and $108$ , which add up to $22 + 65 + 108 = \boxed{195}$ .

Since $1= 15\times23 - 8\times43$ , we need to solve $\begin{aligned} 23x + 43y \; = \; 3000 & = \; 45000 \times 23 - 24000 \times 43 \\ 23(x - 45000) + 43(y + 24000) & = \; 0 \end{aligned}$ Hence $43$ divides $x - 45000$ and so $x \; = \; 45000 - 43n \hspace{1cm} y \; = \; 23n - 24000 \hspace{2cm} n \in \mathbb{Z}$ Since we want both $x$ and $y$ to be positive integers, we need $\tfrac{24000}{23} < n < \tfrac{45000}{43}$ , and hence $1044 \le n \le 1046$ . Thus the possible solutions pairs are $(108,12)$ , $(65,35)$ and $(22,58)$ , making the answer $108 + 65 + 22 = \boxed{195}$ .