Solution:

The conditions say $N = \frac{78}{100}a = \frac{39}{50}a$ and $N = \frac{116}{100}b = \frac{29}{25}b$ for some integers $a$ and $b.$ Since $\gcd(39, 50) = 1,$ the first equation forces $50 \mid a,$ so $N$ is a multiple of $39;$ since $\gcd(29, 25) = 1,$ the second forces $25 \mid b,$ so $N$ is a multiple of $29.$

The least positive integer divisible by both is $N = 39 \cdot 29 = 1131,$ achieved with $a = 1450$ and $b = 975.$ The remainder upon division by $1000$ is $131.$