When $126$ is added to its reversal, $621,$ the sum is $126 + 621 = 747.$ Find the greatest integer which when added to its reversal yields $1211.$
1. We are given that the sum of a number and its reversal is \(1211\). We need to find the greatest integer that satisfies this condition.
2. Let's denote the number as \(100a + 10b + c\), where \(a, b, c\) are digits and \(a \neq 0\) since it is a three-digit number.
3. The reversal of this number is \(100c + 10b + a\).
4. Adding the number and its reversal, we get:
   \[
   (100a + 10b + c) + (100c + 10b + a) = 1211
   \]
5. Simplifying the equation, we have:
   \[
   100a + 10b + c + 100c + 10b + a = 1211
   \]
   \[
   101a + 20b + 101c = 1211
   \]
6. Factoring out common terms, we get:
   \[
   101(a + c) + 20b = 1211
   \]
7. To solve for \(a, b, c\), we need \(101(a + c) + 20b = 1211\). Since \(101(a + c)\) must be a multiple of 101, we can isolate \(a + c\):
   \[
   101(a + c) = 1211 - 20b
   \]
8. For \(101(a + c)\) to be an integer, \(1211 - 20b\) must be divisible by 101. We test values of \(b\) to find a suitable \(a\) and \(c\):
   - If \(b = 5\):
     \[
     1211 - 20 \times 5 = 1211 - 100 = 1111
     \]
     \[
     101(a + c) = 1111
     \]
     \[
     a + c = \frac{1111}{101} = 11
     \]
9. Since \(a + c = 11\), we need to maximize \(a\). The largest possible value for \(a\) is 9, which gives \(c = 2\).
10. Therefore, the number is \(100a + 10b + c = 100 \times 9 + 10 \times 5 + 2 = 900 + 50 + 2 = 952\).

The final answer is \(\boxed{952}\).