Someones age is equal to the sum of the digits of his year of birth. How old is he and when was he born, if it is known that he is older than $11$.
P.s. the current year in the problem is $2010$.
1. Let the year of birth be represented as \(1000a + 100b + 10c + d\), where \(a, b, c, d\) are the digits of the year.
2. Let the current year be \(2010\).
3. The person's age is the sum of the digits of their year of birth, i.e., \(a + b + c + d\).
4. Given that the person is older than \(11\), we have \(a + b + c + d > 11\).

### Case 1: The person has had their birthday this year (2010)
5. The equation for the year of birth and age is:
   \[
   (1000a + 100b + 10c + d) + (a + b + c + d) = 2010
   \]
6. Simplifying, we get:
   \[
   1001a + 101b + 11c + 2d = 2010
   \]
7. Since \(a\) must be \(1\) (as the year is in the 1900s), we substitute \(a = 1\):
   \[
   1001 + 101b + 11c + 2d = 2010
   \]
   \[
   101b + 11c + 2d = 1009
   \]
8. To find \(b\), \(c\), and \(d\), we start with \(b\):
   \[
   101b \leq 1009 \implies b \leq \frac{1009}{101} \approx 9.99 \implies b = 9
   \]
9. Substituting \(b = 9\):
   \[
   101 \cdot 9 + 11c + 2d = 1009
   \]
   \[
   909 + 11c + 2d = 1009
   \]
   \[
   11c + 2d = 100
   \]
10. To find \(c\) and \(d\), we start with \(c\):
    \[
    11c \leq 100 \implies c \leq \frac{100}{11} \approx 9.09 \implies c = 8
    \]
11. Substituting \(c = 8\):
    \[
    11 \cdot 8 + 2d = 100
    \]
    \[
    88 + 2d = 100
    \]
    \[
    2d = 12 \implies d = 6
    \]
12. Therefore, the year of birth is \(1986\) and the age is:
    \[
    2010 - 1986 = 24
    \]

### Case 2: The person has not had their birthday this year (2010)
13. The equation for the year of birth and age is:
    \[
    (1000a + 100b + 10c + d) + (a + b + c + d) = 2009
    \]
14. Simplifying, we get:
    \[
    1001a + 101b + 11c + 2d = 2009
    \]
15. Since \(a\) must be \(1\) (as the year is in the 1900s), we substitute \(a = 1\):
    \[
    1001 + 101b + 11c + 2d = 2009
    \]
    \[
    101b + 11c + 2d = 1008
    \]
16. To find \(b\), \(c\), and \(d\), we start with \(b\):
    \[
    101b \leq 1008 \implies b \leq \frac{1008}{101} \approx 9.98 \implies b = 9
    \]
17. Substituting \(b = 9\):
    \[
    101 \cdot 9 + 11c + 2d = 1008
    \]
    \[
    909 + 11c + 2d = 1008
    \]
    \[
    11c + 2d = 99
    \]
18. To find \(c\) and \(d\), we start with \(c\):
    \[
    11c \leq 99 \implies c \leq \frac{99}{11} = 9 \implies c = 9
    \]
19. Substituting \(c = 9\):
    \[
    11 \cdot 9 + 2d = 99
    \]
    \[
    99 + 2d = 99
    \]
    \[
    2d = 0 \implies d = 0
    \]
20. Therefore, the year of birth is \(1990\) and the age is:
    \[
    2010 - 1990 = 20
    \]

The final answer is \( \boxed{ 24 } \) years old if the person has had their birthday this year, and \(20\) years old if the person has not had their birthday this year.