I am guessing that this is the exact same solution the OP (TheBigOne) got, but I am putting my answer here, so that the thread is answered. Even though this proof requires some non-elementary knowledge (Bertrand's Postulate), it is still a proof. (Although I disagree that Bertrand's Postulate is not elementary. The proof I know is quite elementary, and as i707107 said, Bertrand's Postulate is very widely known.) The solution also utilizes the fact that there are infinitely many prime natural numbers congruent to $2$ modulo $3$. (This fact has an elementary proof.)

First, let $L_k:=\text{lcm}(1,2,\ldots,k)$ for $k=1,2,3,\ldots$. Note that $L_1=1$ and $L_k\geq k(k-1)$ for $k\geq 2$. That is, for each integer $N>0$, $$\sum_{k=1}^N\,\frac{1}{L_k}\leq 1+\sum_{k=2}^N\,\frac{1}{k(k-1)}\leq 1+\sum_{k=2}^\infty\,\frac{1}{k(k-1)}=2\,.$$
This means $S:=\sum\limits_{k=1}^\infty\,\dfrac{1}{L_k}$ exists and is a positive real number less than $2$. (WolframAlpha states that $S\approx 1.77111$.)

We argue by contradiction. Suppose contrary that $S=\dfrac{a}{b}$ for some relatively prime positive integers $a$ and $b$. Let $p_1,p_2,p_3,\ldots$ be the increasing sequence of all prime natural numbers greater than $b$. Using Bertrand's Postulate, $$p_r<p_{r+1}<2\,p_r$$ for all $r=1,2,3,\ldots$. Thus, $$p_{r+1}-p_r\leq p_r-1$$ for each positive integer $r$. Note that, for infinitely many positive integers $r$, it holds that $p_r\equiv 2\pmod{3}$. Therefore, the equality $p_{r+1}-p_r=p_r-1$ does not happen (or else, $p_{r+1}\equiv 0\pmod{3}$). Hence, $p_{r+1}-p_r<p_r-1$ for infinitely many such $r$.

Now,
$$\begin{align}
L_{p_1-1}\,\left(S-\sum_{k=1}^{p_1-1}\,\frac{1}{L_k}\right)&\leq \sum_{r=1}^\infty\,\sum_{k=p_r}^{p_{r+1}-1}\,\frac{L_{p_1-1}}{L_k}\leq \sum_{r=1}^\infty\,\sum_{k=p_r}^{p_{r+1}-1}\,\frac{1}{p_1p_2\cdots p_r}
\\
&=\sum_{r=1}^\infty\,\frac{p_{r+1}-p_r}{p_1p_2\cdots p_r}<\sum_{r=1}^\infty\,\frac{p_{r}-1}{p_1p_2\ldots p_r}
\\
&=\left(1-\frac{1}{p_1}\right)+\left(\frac{1}{p_1}-\frac{1}{p_1p_2}\right)+\left(\frac{1}{p_1p_2}-\frac{1}{p_1p_2p_3}\right)+\ldots
\\
&=1\,.
\end{align}$$
This is a contradiction, as $b\mid L_{p-1}$ and $S>\sum\limits_{k=1}^{p_1-1}\,\frac{1}{L_k}$, which means $L_{p-1}\,\left(S-\sum\limits_{k=1}^{p_1-1}\,\frac{1}{L_k}\right)$ is a positive integer. Therefore, $S$ cannot be a rational number.