Looks extremely tough but is indeed very easy!

Let the number is n.Observe that the remainder is 1 short of the divisor.So if we add 1 to our number n,we see that it is divisible by 2,3,4....10.So to find n+1 find the lcm of 2,3,4...10 and the lcm is 2520.So n+1 is 2520 and n=2519

divided by 2 gives remainder 1, (2 – 1 = 1) divided by 3 gives remainder 2, (3 – 2 = 1) divided by 4 gives remainder 3, (4 – 3 = 1) divided by 5 gives remainder 4, (5 – 4 = 1) divided by 6 gives remainder 5, (6 – 5 = 1) divided by 7 gives remainder 6, (7 – 6 = 1) divided by 8 gives remainder 7, (8 – 7 = 1) divided by 9 gives remainder 8, (9 – 8 = 1) divided by 10 gives remainder 9. (10 – 9 = 1) every time the difference in divisor and remainder is 1

then required number in such case is (LCM divisors – difference)

= (LCM of 2 to 10 ) – 1 = 2520 – 1 = 2519

Easily, one could conclude that each congruence is a generic in $x \equiv -1 (\operatorname{mod} n_i)$ for each $i$ congruence. This means that the solution is LCM (Least Common Multiple) of each $n_i$ in $i$ congruence after subtracted by $1$ as LCM is the least possible integer that satisfy all the congruence. Thus, the answer is, by using python script,

def gcd(a,b):
    while b:
        a, b = b, a%b
    return a

def lcm(a,b):
    return a * b / gcd(a,b)

def lcml(l):
    if len(l) == 1:
        return l[0]
    else:
        return lcm(l[0], lcml(l[1:]))

print lcml([2,3,4,5,6,7,8,9]) - 1

$2519$ .

since the remainders are -1 the divisors that means the next multiple of divisors is +1 the req. number so we will find its LCM and subtract it by 1