Archipelago consists of $ n$ islands : $ I_1,I_2,...,I_n$ and $ a_1,a_2,...,a_n$ - number of the roads on each island.  $ a_1 \equal{} 55$, $ a_k \equal{} a_{k \minus{} 1} \plus{} (k \minus{} 1)$, ($ k \equal{} 2,3,...,n$)
a) Does there exist an island with 2008 roads?
b) Calculate $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n.$
a) We need to determine if there exists an island with 2008 roads. Given the recurrence relation for the number of roads on each island:
\[ a_1 = 55 \]
\[ a_k = a_{k-1} + (k-1) \quad \text{for} \quad k = 2, 3, \ldots, n \]

We can express \( a_n \) in a closed form. Let's prove by induction that:
\[ a_n = 55 + \binom{n}{2} \]

**Base Case:**
For \( n = 1 \):
\[ a_1 = 55 \]
\[ 55 = 55 + \binom{1}{2} \]
\[ 55 = 55 + 0 \]
\[ 55 = 55 \]
The base case holds.

**Inductive Step:**
Assume \( a_k = 55 + \binom{k}{2} \) for some \( k \geq 1 \). We need to show that \( a_{k+1} = 55 + \binom{k+1}{2} \).

Starting from the recurrence relation:
\[ a_{k+1} = a_k + k \]
Using the inductive hypothesis:
\[ a_{k+1} = 55 + \binom{k}{2} + k \]
\[ a_{k+1} = 55 + \frac{k(k-1)}{2} + k \]
\[ a_{k+1} = 55 + \frac{k^2 - k + 2k}{2} \]
\[ a_{k+1} = 55 + \frac{k^2 + k}{2} \]
\[ a_{k+1} = 55 + \frac{k(k+1)}{2} \]
\[ a_{k+1} = 55 + \binom{k+1}{2} \]

Thus, by induction, the formula \( a_n = 55 + \binom{n}{2} \) holds for all \( n \).

To find if there exists an island with 2008 roads:
\[ 2008 = 55 + \binom{n}{2} \]
\[ 2008 - 55 = \binom{n}{2} \]
\[ 1953 = \frac{n(n-1)}{2} \]
\[ 3906 = n(n-1) \]
\[ n^2 - n - 3906 = 0 \]

Solving this quadratic equation:
\[ n = \frac{1 \pm \sqrt{1 + 4 \cdot 3906}}{2} \]
\[ n = \frac{1 \pm \sqrt{15625}}{2} \]
\[ n = \frac{1 \pm 125}{2} \]

This gives us two solutions:
\[ n = \frac{126}{2} = 63 \]
\[ n = \frac{-124}{2} = -62 \]

Since \( n \) must be a positive integer, we have \( n = 63 \). Therefore, there exists an island with 2008 roads.

b) To calculate the sum \( a_1 + a_2 + \cdots + a_n \):
\[ a_1 + a_2 + \cdots + a_n = 55 + \left(55 + \binom{2}{2}\right) + \left(55 + \binom{3}{2}\right) + \cdots + \left(55 + \binom{n}{2}\right) \]
\[ = 55n + \sum_{k=2}^{n} \binom{k}{2} \]

Using the Hockey-Stick Identity:
\[ \sum_{k=1}^{n} \binom{k}{2} = \binom{n+1}{3} \]

Thus:
\[ \sum_{k=2}^{n} \binom{k}{2} = \binom{n+1}{3} - \binom{1}{2} \]
\[ \sum_{k=2}^{n} \binom{k}{2} = \binom{n+1}{3} - 0 \]
\[ \sum_{k=2}^{n} \binom{k}{2} = \binom{n+1}{3} \]

Therefore:
\[ a_1 + a_2 + \cdots + a_n = 55n + \binom{n+1}{3} \]

b) The final answer is \( \boxed{ 55n + \binom{n+1}{3} } \)