A checkerboard is $91$ squares long and $28$ squares wide. A line connecting two opposite vertices of the checkerboard is drawn. How many squares does the line pass through?
To solve this problem, we can use a well-known result in combinatorial geometry. The number of unit squares a diagonal line passes through in a grid of size \( m \times n \) is given by the formula:

\[
m + n - \gcd(m, n)
\]

where \(\gcd(m, n)\) is the greatest common divisor of \(m\) and \(n\).

1. **Identify the dimensions of the checkerboard:**
   The checkerboard is \(91\) squares long and \(28\) squares wide. Thus, \(m = 91\) and \(n = 28\).

2. **Calculate the greatest common divisor (gcd):**
   We need to find \(\gcd(91, 28)\). Using the Euclidean algorithm:
   \[
   \gcd(91, 28) = \gcd(28, 91 \mod 28) = \gcd(28, 7) = \gcd(7, 28 \mod 7) = \gcd(7, 0) = 7
   \]

3. **Apply the formula:**
   Substitute \(m = 91\), \(n = 28\), and \(\gcd(91, 28) = 7\) into the formula:
   \[
   91 + 28 - \gcd(91, 28) = 91 + 28 - 7 = 112
   \]

Thus, the line connecting two opposite vertices of the checkerboard passes through \(112\) squares.

The final answer is \(\boxed{112}\)