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Let $x$ be the number of times the first machine added candy and $y$ be the number of times the second machine removed candy.
From the problem, we have the equation: $5x - 3y = 1000 - 1 = 999$.
To minimize $x$, we want to make $y$ as large as possible. Since the second machine removes $3$ candies each time, $y$ must be a multiple of $3$. The smallest multiple of $3$ that is greater than $999$ is $1002$.
Therefore, $5x - 3(1002) = 999 \implies 5x - 3006 = 999 \implies 5x = 4005 \implies x = \boxed{801}$.
Therefore, the smallest possible number of times the first machine added candy to the vat is $801$.