The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is:

We are asked to add every two-digit positive integer which, on division by 7, leaves a remainder either 2 or 5. In modular notation we are looking for all numbers $$N$$ such that $$N \equiv 2 \pmod{7}$$ or $$N \equiv 5 \pmod{7},$$ with the additional restriction $$10 \le N \le 99.$$

First we treat the remainder 2 case.

A number can be written in the form $$N = 7k + 2$$ where $$k$$ is an integer. We need $$N$$ to be two-digit, so we write

$$10 \le 7k + 2 \le 99.$$

Subtracting 2 everywhere gives

$$8 \le 7k \le 97.$$

Dividing every part by 7 yields

$$\frac{8}{7} \le k \le \frac{97}{7}.$$

The smallest integer $$k$$ satisfying this is $$k = 2,$$ and the largest is $$k = 13.$$ Substituting these back gives the sequence

$$7(2)+2 = 16,\; 7(3)+2 = 23,\; 30,\; 37,\; 44,\; 51,\; 58,\; 65,\; 72,\; 79,\; 86,\; 93.$$

Thus the numbers congruent to 2 (mod 7) are

$$16,\,23,\,30,\,37,\,44,\,51,\,58,\,65,\,72,\,79,\,86,\,93.$$

This is clearly an arithmetic progression with common difference $$d = 7,$$ first term $$a_1 = 16,$$ and last term $$a_{12} = 93.$$ The count of terms is $$n = 12.$$

For the sum of an arithmetic progression we may use the formula

$$S_n = \frac{n}{2}\,(a_1 + a_n).$$

Substituting the known values we obtain

$$S_1 = \frac{12}{2}\,(16 + 93) = 6 \times 109 = 654.$$

Next we consider the remainder 5 case.

Now a number can be written $$N = 7k + 5$$ and must satisfy

$$10 \le 7k + 5 \le 99.$$

Subtracting 5 gives

$$5 \le 7k \le 94.$$

Dividing by 7 gives

$$\frac{5}{7} \le k \le \frac{94}{7}.$$

So the smallest integer $$k$$ is $$k = 1,$$ and the largest is $$k = 13.$$ Plugging in these values lists the terms

$$7(1)+5 = 12,\; 19,\; 26,\; 33,\; 40,\; 47,\; 54,\; 61,\; 68,\; 75,\; 82,\; 89,\; 96.$$

The sequence is

$$12,\,19,\,26,\,33,\,40,\,47,\,54,\,61,\,68,\,75,\,82,\,89,\,96.$$

Again we have an arithmetic progression with $$d = 7,$$ first term $$b_1 = 12,$$ last term $$b_{13} = 96,$$ and $$n = 13$$ terms. Using the same sum formula,

$$S_2 = \frac{13}{2}\,(12 + 96) = \frac{13}{2}\times 108 = 13 \times 54 = 702.$$

Now we combine both sums to get the required total:

$$S = S_1 + S_2 = 654 + 702 = 1356.$$

Hence, the correct answer is Option A.