The positive reals $x_1, x_2, ... , x_n$ have harmonic mean $1$. Find the smallest possible value of $x_1 + \frac{x_2 ^2}{2} + \frac{x_3 ^3}{3} + ... + \frac{x_n ^n}{n}$.
1. **Given Information and Initial Setup:**
   We are given that the positive reals \( x_1, x_2, \ldots, x_n \) have a harmonic mean of 1. The harmonic mean of \( x_1, x_2, \ldots, x_n \) is given by:
   \[
   \frac{n}{\sum_{i=1}^n \frac{1}{x_i}} = 1
   \]
   This implies:
   \[
   \sum_{i=1}^n \frac{1}{x_i} = n
   \]

2. **Expression to Minimize:**
   We need to find the smallest possible value of:
   \[
   S = x_1 + \frac{x_2^2}{2} + \frac{x_3^3}{3} + \cdots + \frac{x_n^n}{n}
   \]

3. **Initial Value Calculation:**
   Let's consider the case where \( x_1 = x_2 = \cdots = x_n = 1 \). In this case:
   \[
   S = 1 + \frac{1^2}{2} + \frac{1^3}{3} + \cdots + \frac{1^n}{n} = \sum_{k=1}^n \frac{1}{k}
   \]
   This sum is known as the Harmonic Series, denoted by \( H_n \).

4. **Lower Bound Proof:**
   We need to show that \( S \geq H_n \) for any positive reals \( x_1, x_2, \ldots, x_n \) with harmonic mean 1.

5. **Inequality Derivation:**
   Consider the term \( \frac{x_i^i}{i} \). We can use the AM-GM inequality to derive a useful inequality:
   \[
   \frac{x_i^i}{i} + \underbrace{\frac{1}{i} + \frac{1}{i} + \cdots + \frac{1}{i}}_{i-1 \text{ times}} \geq x_i
   \]
   This simplifies to:
   \[
   \frac{x_i^i}{i} + \frac{i-1}{i} \geq x_i
   \]
   Rearranging, we get:
   \[
   \frac{x_i^i}{i} \geq x_i - \frac{i-1}{i}
   \]

6. **Summing the Inequalities:**
   Summing this inequality over all \( i \) from 1 to \( n \):
   \[
   \sum_{i=1}^n \frac{x_i^i}{i} \geq \sum_{i=1}^n \left( x_i - \frac{i-1}{i} \right)
   \]
   Simplifying the right-hand side:
   \[
   \sum_{i=1}^n x_i - \sum_{i=1}^n \frac{i-1}{i}
   \]

7. **Using Harmonic Mean Condition:**
   Since the harmonic mean of \( x_1, x_2, \ldots, x_n \) is 1, we have:
   \[
   \sum_{i=1}^n \frac{1}{x_i} = n
   \]
   By the AM-HM inequality:
   \[
   \frac{\sum_{i=1}^n x_i}{n} \geq \frac{n}{\sum_{i=1}^n \frac{1}{x_i}} = 1 \implies \sum_{i=1}^n x_i \geq n
   \]

8. **Combining Results:**
   Therefore:
   \[
   \sum_{i=1}^n \frac{x_i^i}{i} \geq \sum_{i=1}^n x_i - \sum_{i=1}^n \frac{i-1}{i}
   \]
   The second sum is:
   \[
   \sum_{i=1}^n \frac{i-1}{i} = \sum_{i=1}^n 1 - \sum_{i=1}^n \frac{1}{i} = n - H_n
   \]
   Thus:
   \[
   \sum_{i=1}^n \frac{x_i^i}{i} \geq n - (n - H_n) = H_n
   \]

9. **Conclusion:**
   We have shown that:
   \[
   S \geq H_n
   \]
   and the equality holds when \( x_1 = x_2 = \cdots = x_n = 1 \).

The final answer is \( \boxed{H_n} \)