Solution:

Consider any set $S$ representable as the union of $n$ disjoint pairs of integers such that, if any negative number $-m$ is in $S$, then the other element of its pair (henceforth called its "partner") is $\geq 13 m / 6$. Any such $S$ will be called "plausible" of order $n$. We will work with plausible sets (and their partner decompositions) in general; later, we will show how these are relevant to the original problem.
Let $S$ be plausible of order $n$. Then it consists of $n$ disjoint pairs, each of which has a nonnegative sum. Hence, the sum of the elements of $S$ is a nonnegative integer. Now, if we consider the set of all possible values for the sum of $S$, this set consists of nonnegative integers, and it is nonempty (since at least one plausible set exists). Hence, it has a minimal element. So we can meaningfully attempt to characterize a plausible set $S$ (of order $n$) with minimal sum.
This set consists of some negative numbers $-m_{1},-m_{2}, \ldots,-m_{k}$, their (positive) partners, and $2(n-k)$ other nonnegative numbers, all distinct. Clearly, for any fixed choice of the $m_{i}$ and their partners, the sum is minimized by choosing the extra numbers to be the $2(n-k)$ smallest nonnegative integers not equal to the partner of any $-m_{i}$, so we may assume the set to be of this form. Now define $f(m)=\lceil 13 m / 6\rceil$ for any integer $m$. We claim it is no loss to assume the partner of $-m_{i}$ is $f\left(m_{i}\right)$ for each $i$. Indeed, if not, then this partner is $p>f\left(m_{i}\right)$. If $f\left(m_{i}\right) \notin S$, then replacing $p$ by $f\left(m_{i}\right)$ gives us a new plausible set with smaller sum than before, contradicting minimality. So $f\left(m_{i}\right) \in S$. But then $f\left(m_{i}\right)$ has some partner $q$. We claim we can switch partnerships, matching $-m_{i}$ with $f\left(m_{i}\right)$ and $p$ with $q$, so that our partnership decomposition remains plausible. Indeed, if $q \geq 0$, this is clear. If $q<0$ then $q=$ some $-m_{j} \Rightarrow f\left(m_{i}\right) \geq f\left(m_{j}\right)$ since $f\left(m_{i}\right)$ is the partner of $-m_{j}$, but $p>f\left(m_{i}\right) \geq f\left(m_{j}\right)$, so we are again safe. Moreover, in either case, neither of the pairs involved is initially of the form $-m_{l}, f\left(m_{l}\right)$ (in the case $q=-m_{j}$, we cannot have $f\left(m_{i}\right)=f\left(m_{j}\right)$ because $f$ is injective), and after the switch, one of them is of this form, so the total number of such pairs strictly increases. So we can perform this switch as many times as necessary, and eventually, we have each $-m_{i}$ partnered with $f\left(m_{i}\right)$.
Next, we claim that, in the context of the partnering result obtained in the previous paragraph, $-j \in S$ implies $-i \in S$ for $0<i<j$. Indeed, suppose a counterexample exists. If $f(i) \notin S$, then we can replace $-j$ and its partner by $-i$ and $f(i)$ to obtain a new plausible set $S$ whose sum has been changed by $\leq-i+f(i)+j-f(j)=\lceil 7 i / 6\rceil-\lceil 7 j / 6\rceil<0$, contradicting minimality. So we do have $f(i) \in S$. It cannot be partnered with a negative number, since this number would have to be $-i$ (by injectivity of $f)$ but $-i \notin S$; so it is partnered with some positive number. Then, changing this number to $-i$ would again give us a new plausible set with strictly smaller sum, contradiction. This proves the claim.
But this claim shows that the negative elements of $S$ must be exactly $-1,-2, \ldots,-k$, and now we are in a position to estimate the sum of $S$. $S$ consists of these negative elements, their partners $f(1), \ldots, f(k)$, and $2(n-k)$ other, distinct "leftover" nonnegatives which are as small as possible. Let $a$ be the largest leftover, so $0, \ldots, a$ are all in $S$. We claim $b=\lfloor 6 a / 13\rfloor \leq k$; if not, then $-b \notin S$ but $f(b) \leq a$ is in $S$, so it is a leftover (because if its partner is $-c$, injectivity implies $c=b$, impossible), and replacing its nonnegative partner by $-b$ gives a new plausible set with smaller sum, contradiction. So $k \geq b$, and it follows that fully $b$ of the numbers $0, \ldots, a$ are partners of negative numbers. Hence, the number of leftovers can be counted in two ways: $a+1-b=2(n-k)$. But $b \geq 6 a / 13-1 \Rightarrow 2(n-k) \leq a-6 a / 13+2=7 a / 13+2 \Rightarrow a \geq 26(n-k-1) / 7$. Now, the elements of $S$ consist of $-k, \ldots,-1,0, \ldots, a$, and the partners of $-k, \ldots,-(b+1)$. So the sum of these numbers is
$$\frac{a(a+1)}{2}-\frac{k(k+1)}{2}+\sum_{i=b+1}^{k}\lceil 13 i / 6\rceil \geq \frac{a(a+1)}{2}-\frac{k(k+1)}{2}+\frac{13}{6}\left(\frac{k(k+1)}{2}-\frac{b(b+1)}{2}\right) ;$$
using the (rather crude) estimates $0 \leq a, b, k \leq 2 n$ and $b \leq 6 a / 13$ to simplify our calculations, this is
$$\begin{gathered} \geq \frac{a^{2}}{2}-\frac{k^{2}}{2}+\frac{13 k^{2}}{12}-\frac{13 b^{2}}{12}-4 n \geq \frac{a^{2}}{2}+\frac{7 k^{2}}{12}-\frac{13(6 a / 13)^{2}}{12}-4 n=\frac{7 a^{2}}{26}+\frac{7 k^{2}}{12}-4 n \\ \geq \frac{26(n-k-1)^{2}}{7}+\frac{7 k^{2}}{12}-4 n \geq \frac{26(n-k)^{2}}{7}-8 n+\frac{7 k^{2}}{12}-4 n=\frac{361 k^{2}}{84}-\frac{52 n k}{7}+\frac{26 n^{2}}{7}-12 n \end{gathered}$$
$$=\frac{361}{84}\left(k-\frac{312 n}{361}\right)^{2}+\frac{1274}{2527} n^{2}-12 n \geq \alpha n^{2}-12 n$$
where $\alpha=1274 / 2527>1 / 2$. So we now know that any plausible set of order $n$ has sum $\geq \alpha n^{2}-12 n$.
Now let us return to the original problem statement. Suppose that the statement is false; we seek a contradiction. Let $i_{1}<i_{2}<\cdots<i_{c}$ be the (finitely many) values of $i$ for which $S_{i}$ has an element $>13 i / 7$, and let $d=i_{1}+i_{2}+\cdots+i_{c}$. Now suppose $i$ is not equal to any of $i_{1}, \ldots, i_{c}$, and suppose $-m \in S_{i}(m>0)$. Then the other element of $S_{i}$ is $i+m$, and we have $i+m \leq$ $13 i / 7 \Rightarrow m \leq 6 i / 7 \Rightarrow i+m \geq 13 m / 6$. Consequently, if we take any $n>i_{c}-c$, we see that the set $S=\left(S_{1} \cup S_{2} \cup \cdots \cup S_{n+c}\right)-\left(S_{i_{1}} \cup S_{i_{2}} \cup \cdots \cup S_{i_{c}}\right)$ is plausible of order $n$. So the sum of its elements is $\geq \alpha n^{2}-12 n$. However, this sum should also equal
$$(1+2+\cdots+(n+c))-\left(i_{1}+i_{2}+\cdots+i_{c}\right)=(n+c)(n+c+1) / 2-d$$
This is a quadratic function of $n$ with leading coefficient $1 / 2$, so it will actually be less than $\alpha n^{2}-12 n$ when $n$ is sufficiently large. This is our contradiction.