11. In a school's spring sports meeting, several students form a rectangular formation with 8 columns. If 120 more students are added to the original formation, a square formation can be formed; if 120 students are removed from the original formation, a square formation can also be formed. How many students are there in the original rectangular formation?
11. Let the original rectangular formation have $8x$ students. From the given conditions, we know that $8x+120$ and $8x-120$ are both perfect squares. Therefore, we can set
$$
\left\{\begin{array}{l}
8x+120=m^{2}, \\
8x-120=n^{2}.
\end{array}\right.
$$

where $m$ and $n$ are positive integers, and $m > n$.
Subtracting (2) from (1), we get $m^{2}-n^{2}=240$, which is
$$
(m+n)(m-n)=240=2^{4} \times 3 \times 5.
$$

From equations (1) and (2), we know that $m^{2}$ and $n^{2}$ are both multiples of 8, so $m$ and $n$ are both divisible by 4. Therefore, $m+n$ and $m-n$ are both divisible by 4, leading to
$$
\left\{\begin{array}{l}
m+n=60, \\
m-n=4
\end{array} \text { or } \left\{\begin{array}{l}
m+n=20, \\
m-n=12.
\end{array}\right.\right.
$$

Solving these, we get $\left\{\begin{array}{l}m=32, \\ n=28\end{array}\right.$ or $\left\{\begin{array}{l}m=16, \\ n=4.\end{array}\right.$
Thus, $8x=m^{2}-120=32^{2}-120=904$, or $8x=m^{2}-120=16^{2}$
$$
-120=136.
$$

Therefore, the original rectangular formation has 136 students or 904 students.