Variant of the Strong Law of Large Numbers

Yes. Here is a proof sketch:

1) The statement is equivalent to the statement $\frac{Z_n}{n}\rightarrow 2$ with prob 1, where $$ Z_n = \sum_{i=1}^n \sum_{j \in \{1, ..., n\}, j \neq i} 1_{\{|X_i-X_j|\leq 1/n\}}$$

2) We get $E[\frac{Z_n}{n}]\rightarrow 2$ and $Var(\frac{Z_n}{n}) = O(1/n)$. Thus, $Var(\frac{Z_{n^2}}{n^2})=O(1/n^2)$ and hence $$ \frac{Z_{n^2}}{n^2} \rightarrow 2 \quad \mbox{with prob 1} $$

3) For indices $k$ such that $n^2\leq k <(n+1)^2$, unfortunately we cannot "quite" say that $\frac{Z_{n^2}}{(n+1)^2} \leq \frac{Z_k}{k} \leq \frac{Z_{(n+1)^2}}{n^2}$ because the indicators $1_{\{|X_i-X_j|\leq 1/n\}}$ now have dependence on $n$. So we have to go to step 4:

---

Here is a fix:

4) Let $\{a_n\}_{n=1}^{\infty}$ be a deterministic sequence of (possibly negative) integers that satisfies $a_n = O(\sqrt{n})$ and $n + a_n \in \{1, 2, 3, \ldots\}$ for all positive integers $n$. Define: $$R_n(\{a_n\}) := \sum_{i=1}^n \sum_{j \in \{1, …, n\}, j \neq i} 1_{\{|X_i-X_j|\leq 1/(n+a_n)\}}$$ Now we can equally say $E[\frac{R_n(\{a_n\})}{n}]\rightarrow 2$ and $Var(\frac{R_n(\{a_n\})}{n})=O(1/n)$, and so $R_{n^2}(\{a_n\})/n^2\rightarrow 2$ with prob 1. Furthermore for any $k$ such that $n^2\leq k <(n+1)^2$ we get $$ \frac{R_{n^2}(\{a_n\})}{(n+1)^2}\leq\frac{Z_k}{k} \leq \frac{R_{(n+1)^2}(\{b_n\})}{n^2}$$ for some sequences $\{a_n\}$ and $\{b_n\}$.