We claim that there do exist such positive integers. Notice: $\sum_{i=1}^{n} \frac{a_i-b_i}{a_i+b_i} = \sum_{i=1}^{n} \left(1 - \frac{2b_i}{a_i+b_i}\right) = n - 2 \sum_{i=1}^{n} \frac{b_i}{a_i+b_i}$ It then suffices to find such $a_i$, $b_j$ so that $\frac{1}{2} < \sum_{i=1}^{n} \frac{b_i}{a_i+b_i} < \frac{1}{2} + \frac{1}{2 \cdot 1998}$. We will prove a stronger result by choosing each element as a function of a parameter $k$ so that for each $\varepsilon > 0$, there exists a positive integral choice of $k$ such that: $\frac{1}{2} < \sum_{i=1}^{n} \frac{b_i(k)}{a_i(k)+b_i(k)} < \frac{1}{2} + \varepsilon$ Take: Let $m = \frac{(n-1)(n-2)}{2}k$ and choose $a_1 = m(k-2), a_2 = k, a_3 = 2k, \ldots, a_{n-1} = (n-2)k$ and $b_1 = m(k+2), b_2 = 1, b_3 = 2, \ldots, b_{n-1} = n-2, b_n = n-1$. To achieve the equal sums condition, choose $a_n = 4m + \frac{n(n-1)}{2} - k \frac{(n-2)(n-1)}{2}$. Furthermore, choose $k$ large enough so that any two elements are distinct from each other. Then: $\sum_{i=1}^{n} \frac{b_i}{a_i+b_i} = \frac{1}{2} + \frac{1}{k} + \frac{n-2}{k+1} + \frac{1}{\frac{3}{2}(n-2)k + \frac{n}{2} + 1}$ For $n \ge 3$, it is clear that this sum is always larger than $\frac{1}{2}$ and approaches $\frac{1}{2}$ from above as $k \to \infty$ so we can find sufficient $k$ for any $\varepsilon > 0$. For $n =2$, we use a similar approach with $a_1 = k^2-2k$, $a_2 = 1+4k$, $b_1 = k^2+2k$, and $b_2 = 1$ (for sufficiently large $k$ so that any pair of elements are distinct).