To find the sum of the given series, we can first simplify the terms by finding a common denominator:

1/3 + 1 + 5/3 + 7/3 + 3 + ... + 101/3 + 103/3 + 35

The common denominator for all the terms is 3. Now, let's rewrite the series:

(1/3 + 5/3 + 7/3 + ... + 101/3 + 103/3) + (1 + 3 + ... + 35)

The first part of the series is an arithmetic series with a common difference of 2. We can use the formula for the sum of an arithmetic series to find its value:

Sum = (n/2)(first term + last term)
= (n/2)(2a + (n-1)d)

In this case, the first term (a) is 1/3, the last term is 103/3, and the common difference (d) is 2. Let's calculate the number of terms (n):

103/3 = 1/3 + (n-1)(2)
103/3 = 1/3 + 2n - 2
2n = 103/3 + 2 - 2/3
2n = 103/3 + 6/3 - 2/3
2n = 107/3
n = (107/3) / 2
n = 107/6

Now, let's substitute these values into the formula for the sum of an arithmetic series:

Sum = (n/2)(2a + (n-1)d)
= (107/6 / 2)(2(1/3) + ((107/6)-1)(2))
= (107/12)(2/3 + (107/6 - 1)(2))
= (107/12)(2/3 + (107/6 - 6/6)(2))
= (107/12)(2/3 + (101/6)(2))
= (107/12)(2/3 + 202/6)
= (107/12)(2/3 + 101/3)
= (107/12)(103/3)
= (107 * 103) / (12 * 3)
= 11021 / 36

Now, let's calculate the second part of the series, which is also an arithmetic series:

Sum = (n/2)(first term + last term)
= (35/2)(1 + 35)
= (35/2)(36)
= 630

Finally, let's add the two parts of the series together:

11021/36 + 630 = (11021 + 22680) / 36
= 33601 / 36

Therefore, the sum of the given series is 33601/36.