1000 chocolates? Yummy!

Sandeep eats the chocolates labelled $3 , 6 , 9 , 12 , \cdots , 999$ . It ends in $999$ because it is the last number less than $1000$ which is divisible by $3$ .
Abhiram eats the chocolates labelled $7 , 14 , 21 , 28 , \cdots , 994$ . It ends in $994$ because it is the last number less than $1000$ which is divisible by $7$ .

Let $999$ be the nth chocolate in the series $3 , 6 , 9, \cdots , 999$ .
So, $999 = 3 + (n - 1)×3\\ n = 333$ .

Let $994$ be the mth chocolate in the series $7 , 14 , 21, \cdots , 994$ .
So, $994 = 7 + (m - 1)×7\\ m = 142$ .

The chocolates which are eaten by both is the progression formed by the multiples of $lcm(3 , 7) = 21$ . So, the progression is $21, 42, 63, \cdots, 987$ . It ends in $987$ because it is the last number before $1000$ which is divisible by $21$ . Let $987$ be the pth number in the given series.
So, $987 = 21 + (p - 1)×21\\ p = 47$ .

So, the number of chocolates that are left over = $1000 - \left(\left(n + m\right) - p\right)\\ = 1000 - \left(\left(333 + 142\right) - 47\right)\\ = 1000 - \left(475 - 47\right)\\ = 1000 - 428\\ = \color{#69047E}{\boxed{572}}$ .