You can use PROC OPTMODEL in SAS/OR to formulate the problem and call either the MILP solver or possibly the CLP solver:
SAS/OR(R) 13.2 User's Guide: Mathematical Programming
Besides the v and w decision variables, you need a binary decision variable x[j,k] that indicates whether observation j fails constraint k. You want to fix x[j,k] = 0 for all observations j with FLAG = 0 and all constraints k. If you are using the MILP solver, you can enforce this relationship by introducing big-M constraints (assuming DC also depends on j):
/* if x[j,k] = 1 then v + sum {i in ISET} w * DC[i,j] >= 0 */
-v - sum {i in ISET} w * DC[i,j] <= M1[j,k] * (1 - x[j,k])
/* if x[j,k] = 0 then v + sum {i in ISET} w * DC[i,j] < 0 */
v + sum {i in ISET} w * DC[i,j] + epsilon <= M2[j,k] * x[j,k]
where M1[j,k] and M2[j,k] are upper bounds on the left hand sides, and epsilon is a small positive number, say 1e-6, that represents how far away from 0 you consider "negative" to be.
If you are using the CLP solver, you can enforce the relationship with a single set of REIFY constraints, without big-M or epsilon:
REIFY(x[j,k], v + sum {i in ISET} w * DC[i,j] >= 0)
But note that the CLP solver requires all variables to be integer, and I don't know whether your v and w have that restriction.
There is a bit of ambiguity in the problem statement. Does "fail these constraints" mean fail all constraints or fail at least one constraint?
In either case, you need another binary variable y that indicates whether observation j fails, and your objective is to maximize sum {j in JSET} y. If y = 1 means fail all, then you want y <= x[j,k] for all k. If y = 1 means fail at least one, then you want y <= sum {k in KSET} x[j,k].
