\begin{equation*} \begin{split} {240 \choose 120} & = \frac{240!}{120!(240-120)!} = \frac{120! \cdot 121 \cdot 122 \cdot \cdots \cdot 240}{120! \cdot 120!} = \frac{121 \cdot 122 \cdot \cdots \cdot 240}{120!} \\ & = \frac{122 \cdot 124 \cdot 126 \cdot \cdots \cdot 236 \cdot 238 \cdot 240}{\;\,61\cdot\;\,62\cdot\;\,62\cdot \cdots \cdot 117 \cdot 119 \cdot 120} \cdot \frac{121 \cdot 123 \cdot 125 \cdot \cdots \cdot 237 \cdot 239}{60!} \\ & = 2^{60} \cdot \frac{121 \cdot 123 \cdot 125 \cdot \cdots \cdot 237 \cdot 239}{60!} =\frac{2^{60} \cdot 239!!}{60! \cdot 119!!} \end{split} \end{equation*} \begin{equation*} {240 \choose 120} \cdot \left(60! \cdot 119!! \right) = 2^{60} \cdot 239!! \end{equation*} \begin{equation*} 60! \cdot 119!! = \frac{60!}{11 \cdot 2} \cdot \frac{119!!}{11} \cdot 2 \cdot 11 \cdot 11 = \frac{60!}{11 \cdot 2} \cdot \frac{119!!}{11} \cdot 242 \end{equation*} $242 \mid (119!! \cdot 60!)$
$(119!! \cdot 60!) \mid (2^{60} \cdot 239!!)$
By the transitive property, $242 \mid (2^{60} \cdot 239!!)$
\begin{equation*} {240 \choose 120} = 242 \cdot \frac{2^{59} \cdot \frac{239!!}{121}}{60! \cdot 119!!} \end{equation*}
How do I prove that $\frac{2^{59} \cdot \frac{239!!}{121}}{60! \cdot 119!!}$ is an integer?
Worth noting: This is all closely related to the Catalan Numbers. Indeed, the integrality of those at least implies that $121$ divides your number. A cheap (but unintuitive) proof of that can be obtained via the (easily proven) general relation $$\binom {2n}n- \binom {2n}{n+1}=\frac 1{n+1}\binom {2n}n$$ (you would still have to show that your number was even).
Here's a direct proof:
de polignac's formula is very useful for problems like this. It tells us that, for a prime $p$, $$v_p(n!)=\sum_{i=1}^{\infty} \Bigg \lfloor \frac n{p^i}\Bigg \rfloor$$
Where, as usual, $v_p(m)$ denotes the greatest power of $p$ which divides $m$.
Now, you are interested in $$\alpha=\frac {240!}{(120!)^2}$$
Noting that $242=2\times 11^2$ we compute $$v_2(240!)=237\quad \& \quad v_2(120!)=116$$
Since $237>2\times 116$ we see that $\alpha$ is even. Indeed, $$v_2(\alpha)=237-2\times 116=5$$
We also compute $$v_{11}(240!)=22\quad \&\quad v_{11}(120!)=10$$
which implies $$v_{11}(\alpha)=22-2\times 10=2$$ and we are done.