Consider and odd prime $p$. For each $i$ at $\{1, 2,..., p-1\}$, let $r_i$ be the rest of $i^p$ when it is divided by $p^2$.
Find the sum:
$r_1 + r_2 + ... + r_{p-1}$
1. **Understanding the problem**: We need to find the sum of the remainders when \(i^p\) is divided by \(p^2\) for \(i\) ranging from 1 to \(p-1\), where \(p\) is an odd prime.

2. **Using Fermat's Little Theorem**: Fermat's Little Theorem states that for any integer \(a\) and a prime \(p\), \(a^p \equiv a \pmod{p}\). However, we need to consider the expression modulo \(p^2\).

3. **Binomial Expansion**: Consider the binomial expansion of \((p+k)^p\):
   \[
   (p+k)^p = \sum_{j=0}^{p} \binom{p}{j} p^{p-j} k^j
   \]
   Since \(p\) is a prime, all binomial coefficients \(\binom{p}{j}\) for \(1 \leq j \leq p-1\) are divisible by \(p\). Therefore, modulo \(p^2\), we have:
   \[
   (p+k)^p \equiv p^p + k^p \pmod{p^2}
   \]
   Since \(p^p\) is clearly divisible by \(p^2\) (as \(p \geq 2\)), we get:
   \[
   (p+k)^p \equiv k^p \pmod{p^2}
   \]

4. **Symmetry in the Remainders**: Notice that for each \(i\) in \(\{1, 2, \ldots, p-1\}\), the remainder \(r_i\) of \(i^p\) modulo \(p^2\) is the same as the remainder of \((p-i)^p\) modulo \(p^2\). This is because:
   \[
   (p-i)^p \equiv -i^p \pmod{p^2}
   \]
   Since \(p\) is odd, \(-i^p \equiv p^2 - i^p \pmod{p^2}\).

5. **Summing the Remainders**: We need to sum the remainders \(r_i\) for \(i\) from 1 to \(p-1\). Using the symmetry, we can pair the terms:
   \[
   r_1 + r_2 + \cdots + r_{p-1}
   \]
   Each pair \((i, p-i)\) contributes \(p^2\) to the sum. There are \(\frac{p-1}{2}\) such pairs.

6. **Final Calculation**: The sum of the remainders is:
   \[
   \sum_{i=1}^{p-1} r_i = \frac{p-1}{2} \cdot p^2
   \]

The final answer is \(\boxed{\frac{p^2(p-1)}{2}}\).