Let the number be n.
Then:

n must be 1 more than a multiple of 2
n must be 1 more than a multiple of 3
n must be 1 more than a multiple of 4
n must be 1 more than a multiple of 5
n must be 1 more than a multiple of 6
n must be a multiple of 7

So there is some positive integer p such that n = 7p

The least common multiple of 2,3,4,5,6 is 60

So n must be 1 more than a multiple of 60.

So there is some positive integer q such that n = 60q+1

So we have the Diophantine equation:

7p = 60q + 1

Write 60 in terms of its nearest multiple of 7, which is 63

7p = (63-3)q + 1

7p = 63q - 3q + 1

Divide through by 7

p = 9q -3q/7 + 1/7

3q/7 - 1/7 = 9q - p

The right side is an integer so the left side must equal
that same integer.  Suppose that integer is A:

3q/7 - 1/7 = A;  9q - p = A

3q - 1 = 7A

Write 7 in terms of its nearest multiple of 3 which is 6.

3q - 1 = (6+1)A

3q - 1 = 6A + A

Divide through by 3

q - 1/3 = 2A + A/3

q - 2A = A/3 + 1/3

The left side is an integer so the right side must equal
that same integer.  Suppose that integer is B:

A/3 + 1/3 = B;    q - 2A = B

A + 1 = 3B

A = 3B - 1

q - 2A = B

q - 2(3B - 1) = B

q - 6B + 2 = B

q = 7B - 2

9q - p = A

9(7B - 2) - p = 3B - 1 

63B - 18 - p = 3B - 1
60B - 17 = p

Since n = 7p
      n = 7(60B-17)
      n = 420B-119

So the smallest positive value of n is when B=1

      n = 420(1)-119
      n = 301  

Edwin