Let $c_0, c_1, \ldots, c_{24}$ be these 25 4-digit integers in some order let $N$ be the 100-digit number $N = \sum_{k=0}^{24} c_k \cdot 10^{4k}$. The fact that $10^4 \equiv (-1)^4 \equiv 1 \pmod{11}$ give us $N = \sum_{k=0}^{24} a_k \equiv 12 \cdot 2003 + \sum_{m=-1}^{11} (1991 + m) \equiv 1 \cdot 1 + \sum_{m=-1}^{12} m = \sum_{m=1}^{11} m = \frac{11 \cdot 12}{2} = 11 \cdot 6 \equiv 0 \pmod{11}$, i.e. $11 \mid N$. Consequently $N$ is not a prime.