The number of apples decreases by $1$ after $1$ trick of type $B$ or $C$ is performed, and increases by $2$ after $1$ trick of type $A$ is performed. The number of grapes decreases by $1$ after $1$ trick of type $A$ or $B$ is performed, and increases by $4$ after $1$ trick of type $C$ is performed. Suppose the magician used $x$ tricks of type $A$, $y$ tricks of type $B$ and $z$ tricks of type $C$ to reach the situation where the numbers of apples and grapes are the same as their initial numbers, then we must have
$$y+z=2x, \quad x+y=4z.$$
Since $x, y, z$ are non-negative integers and at least one of $x, y, z$ must be positive, we see that the solution of the simultaneous equation above must have the form $(x, y, z) = (5k, 7k, 3k)$, where $k$ is a positive integer. Since the number of oranges decreases by $1$ after $1$ trick of type $A$ or $C$ is performed and increases by $3$ after $1$ trick of type $B$ is performed, the change of the number of oranges after $5k$ tricks of type $A$, $7k$ tricks of type $B$ and $3k$ tricks of type $C$ are performed is given by $-5k+3 \times 7k-3k = 13k$. $k=1$, then, gives the smallest possible number of oranges when the number is $2011$ for both apples and grapes, which is $2011+13=2024$.