Penn writes a 2013-term arithmetic sequence of positive integers, and Teller writes a different 2013-term arithmetic sequence of integers. Teller's first term is the negative of Penn's first term. Each then finds the sum of terms in his sequence. If their sums are equal, then what is the smallest possible values of the first term in Penn's sequence?

To find the smallest possible value of the first term in Penn's sequence, let's break down the problem.

Given:
- Penn writes a 2013-term arithmetic sequence of positive integers.
- Teller writes a different 2013-term arithmetic sequence of integers.
- Teller's first term is the negative of Penn's first term.
- Each finds the sum of terms in his sequence.

Let's assume that the first term of Penn's sequence is 'a', and the common difference is 'd'.
This means that the first term of Teller's sequence is '-a' (negative of Penn's first term), and the common difference is also 'd'.

Now, let's find the sum of terms in Penn's sequence:
Sum of an arithmetic sequence = (Number of terms / 2) * (2 * first term + (Number of terms - 1) * common difference)
Sum of Penn's sequence = (2013 / 2) * (2a + (2013 - 1) * d) = 1006(2a + 2012d)

Similarly, let's find the sum of terms in Teller's sequence:
Sum of Teller's sequence = (2013 / 2) * (2(-a) + (2013 - 1) * d) = 1006(-2a + 2012d)

According to the given condition, the sums of their sequences are equal:
1006(2a + 2012d) = 1006(-2a + 2012d)

Now, let's simplify the equation:
2a + 2012d = -2a + 2012d (canceling out the 1006 factors)

This equation shows that the first term 'a' cancels out, which means the value of the first term doesn't matter. The only thing that matters is the common difference 'd'.

Therefore, there is no smallest possible value for the first term in Penn's sequence, as any positive integer value of 'a' will result in the same sum as long as the common difference remains the same.