Let $ (a_n)^{\infty}_{n\equal{}1}$ be a sequence of integers with $ a_{n} < a_{n\plus{}1}, \quad \forall n \geq 1.$ For all quadruple $ (i,j,k,l)$ of indices such that $ 1 \leq i < j \leq k < l$ and $ i \plus{} l \equal{} j \plus{} k$ we have the inequality $ a_{i} \plus{} a_{l} > a_{j} \plus{} a_{k}.$ Determine the least possible value of $ a_{2008}.$
1. Given the sequence \( (a_n)_{n=1}^{\infty} \) of integers with \( a_n < a_{n+1} \) for all \( n \geq 1 \), we need to determine the least possible value of \( a_{2008} \) under the condition that for all quadruples \( (i, j, k, l) \) such that \( 1 \leq i < j \leq k < l \) and \( i + l = j + k \), the inequality \( a_i + a_l > a_j + a_k \) holds.

2. Consider the inequality \( a_i + a_l > a_j + a_k \) for \( i + l = j + k \). This implies that the sum of the terms at the outer indices is greater than the sum of the terms at the inner indices.

3. To find a lower bound for \( a_n \), we start by examining the sequence properties. We assume \( a_i \) is an increasing sequence of integers. Let's derive a recursive relationship:
   \[
   a_i + a_{i-2} \geq 2a_{i-1} + 1
   \]
   This inequality suggests that the difference between consecutive terms increases as \( i \) increases.

4. We can rewrite the inequality as:
   \[
   a_i - a_{i-1} \geq a_{i-1} - a_{i-2} + 1
   \]
   Let \( d_i = a_i - a_{i-1} \). Then:
   \[
   d_i \geq d_{i-1} + 1
   \]
   This implies that \( d_i \) is an increasing sequence of integers.

5. Starting from \( d_2 = a_2 - a_1 \), we have:
   \[
   d_3 \geq d_2 + 1, \quad d_4 \geq d_3 + 1, \quad \ldots, \quad d_i \geq d_{i-1} + 1
   \]
   Summing these inequalities, we get:
   \[
   d_i \geq d_2 + (i-2)
   \]

6. Assuming \( d_2 = 1 \) (the smallest possible value for an increasing sequence of integers), we have:
   \[
   d_i \geq 1 + (i-2) = i-1
   \]

7. Now, summing the differences \( d_i \) from \( i = 2 \) to \( n \), we get:
   \[
   a_n - a_1 = \sum_{i=2}^n d_i \geq \sum_{i=2}^n (i-1)
   \]
   The sum of the first \( n-1 \) natural numbers is:
   \[
   \sum_{i=1}^{n-1} i = \frac{(n-1)n}{2}
   \]
   Therefore:
   \[
   a_n - a_1 \geq \frac{(n-1)n}{2}
   \]

8. Assuming \( a_1 = 1 \) (the smallest possible value for the first term), we get:
   \[
   a_n \geq 1 + \frac{(n-1)n}{2}
   \]

9. For \( n = 2008 \):
   \[
   a_{2008} \geq 1 + \frac{2007 \cdot 2008}{2} = 1 + 2007 \cdot 1004 = 1 + 2015028 = 2015029
   \]

The final answer is \( \boxed{2015029} \).