A finite [set](https://artofproblemsolving.com/wiki/index.php/Set) $\mathcal{S}$ of distinct real numbers has the following properties: the [mean](https://artofproblemsolving.com/wiki/index.php/Arithmetic_mean) of $\mathcal{S}\cup\{1\}$ is $13$ less than the mean of $\mathcal{S}$, and the mean of $\mathcal{S}\cup\{2001\}$ is $27$ more than the mean of $\mathcal{S}$. Find the mean of $\mathcal{S}$.
Let $x$ be the mean of $\mathcal{S}$. Let $a$ be the number of elements in $\mathcal{S}$.
Then, the given tells us that $\frac{ax+1}{a+1}=x-13$ and $\frac{ax+2001}{a+1}=x+27$. Subtracting, we have 
\begin{align*}\frac{ax+2001}{a+1}-40=\frac{ax+1}{a+1} \Longrightarrow \frac{2000}{a+1}=40 \Longrightarrow a=49\end{align*}
We plug that into our very first formula, and get:
\begin{align*}\frac{49x+1}{50}&=x-13 \\ 49x+1&=50x-650 \\ x&=\boxed{651}.\end{align*}