Using algebra
Let the five consecutive positive integers be x-2, x-1, x, x + 1, x + 2. Their sum is 5x, so 5x = 2005, that is x = 401. The five numbers are therefore 399, 400, 401, 402, 403, so the smallest number is 399.
Using approximation/averages
The numbers are close to 2005\div5 = 401 401+401+401+401+401=2005\\
401+400+401+402+401=2005\\
399+400+401+402+403=2005