Let $a, b, c$ be two-digit, three-digit, and four-digit numbers, respectively. Assume that the sum of all digits of number $a+b$, and the sum of all digits of $b + c$ are all equal to $2$. The largest value of $a + b + c$ is

(A): $1099$ (B): $2099$ (C): $1199$ (D): $2199$ (E): None of the above.
1. **Understanding the problem:**
   - \(a\) is a two-digit number.
   - \(b\) is a three-digit number.
   - \(c\) is a four-digit number.
   - The sum of the digits of \(a + b\) is 2.
   - The sum of the digits of \(b + c\) is 2.
   - We need to find the largest value of \(a + b + c\).

2. **Analyzing the constraints:**
   - The sum of the digits of \(a + b\) is 2. This implies that \(a + b\) must be a number whose digits sum to 2.
   - The sum of the digits of \(b + c\) is 2. This implies that \(b + c\) must be a number whose digits sum to 2.

3. **Maximizing \(a\):**
   - Since \(a\) is a two-digit number, the maximum value of \(a\) is 99. However, we need to ensure that \(a + b\) has digits summing to 2.

4. **Finding \(b\) such that \(a + b\) has digits summing to 2:**
   - If \(a = 99\), then \(b\) must be such that the digits of \(99 + b\) sum to 2.
   - The smallest \(b\) that satisfies this is \(b = 911\) because \(99 + 911 = 1010\) and the digits of 1010 sum to 2.

5. **Maximizing \(c\):**
   - We need to find \(c\) such that \(b + c\) has digits summing to 2.
   - If \(b = 911\), then \(c\) must be such that the digits of \(911 + c\) sum to 2.
   - The smallest \(c\) that satisfies this is \(c = 9189\) because \(911 + 9189 = 10100\) and the digits of 10100 sum to 2.

6. **Calculating \(a + b + c\):**
   - \(a = 99\)
   - \(b = 911\)
   - \(c = 9189\)
   - Therefore, \(a + b + c = 99 + 911 + 9189 = 10199\).

7. **Verifying the solution:**
   - The sum of the digits of \(a + b = 1010\) is \(1 + 0 + 1 + 0 = 2\).
   - The sum of the digits of \(b + c = 10100\) is \(1 + 0 + 1 + 0 + 0 = 2\).
   - Both conditions are satisfied.

The final answer is \(\boxed{10199}\).