Is this problem that's similar to integer linear programming also an NP-complete problem?
Your problem is solvable in polynomial time.

Your problem is equivalent to ILP with difference constraints, i.e., every linear inequality has the form $x_j - x_i \ge c_{i,j}$.  (Why?  Set $x_i = v_1 + \dots + v_i$; then $v_{i+1} + \dots + v_j = x_j - x_i$.)

Such a system of inequalities can be solved with shortest-path algorithms.  In particular, if we didn't have the restriction to integer values, this becomes the standard problem of solving a system of difference constraints.  You can solve this by creating a graph with one vertex per variable and one edge per constraints, and finding shortest paths using Bellman-Ford.  See, e.g., Algorithms: Difference Constraints, Decide whether a set of inequalities is solvable, http://videolectures.net/mit6046jf05_demaine_lec18/, or a good textbook (I think this is described in CLRS).

As stated, those algorithms don't take into account the restriction to integer solutions.  But if you examine how shortest path algorithms work, if all of the constants are integers, then the solution will be an integer solution as well.  So, that ensures you get an integer solution.

Lastly, the requirement that $v_1 + \dots + v_n = c$ is equivalent to $x_n = c$, which can be handled using the shortest paths algorithm.  In particular, find a solution; then if $x_n \ne c$, then set $c' = c - x_n$ and add the constant $c'$ to all variables; the result will be another solution that satisfies all the inequalities, and also satisfies the equality $x_n = c$.

Finally, once you've solved to find a solution for the $x$'s, you can recover a solution for the $v$'s by setting $v_i = x_i - x_{i-1}$.

One catch is that this solution might not be all positive.  To handle this, I suggest adding the inequalities $x_i - x_{i-1} \ge 0$ before solving.  That will then meet all of your needs.