Solution:

Let $\mathcal{S}$ have $n$ elements with mean $x,$ so the elements sum to $nx.$ The two conditions say $$\frac{nx + 1}{n + 1} = x - 13 \qquad \text{and} \qquad \frac{nx + 2001}{n + 1} = x + 27.$$

Clearing denominators, $nx + 1 = (n+1)x - 13(n+1)$ gives $x - 13(n+1) = 1,$ and $nx + 2001 = (n+1)x + 27(n+1)$ gives $x + 27(n+1) = 2001.$ Subtracting the first equation from the second yields $40(n+1) = 2000,$ so $n + 1 = 50.$

Then $x = 1 + 13 \cdot 50 = 651.$