A sequence of numbers $ a_1, a_2, a_3, ...$ satisfies

(i) $ a_1 \equal{} \frac{1}{2}$
(ii) $ a_1\plus{}a_2 \plus{} \cdots \plus{} a_n \equal{} n^2 a_n \ (n \geq 1)$

Determine the value of $ a_n \ (n \geq 1)$.
1. **Base Case:**
   We start by verifying the base case for \( n = 1 \).
   \[
   a_1 = \frac{1}{2} = \frac{1}{1 \cdot 2}
   \]
   This satisfies the given condition.

2. **Inductive Hypothesis:**
   Assume that for some \( k \geq 1 \), the formula holds:
   \[
   a_k = \frac{1}{k(k+1)}
   \]
   We need to show that this implies:
   \[
   a_{k+1} = \frac{1}{(k+1)(k+2)}
   \]

3. **Sum of the Sequence:**
   By the problem's condition, we have:
   \[
   a_1 + a_2 + \cdots + a_k = k^2 a_k
   \]
   Using the inductive hypothesis:
   \[
   a_1 + a_2 + \cdots + a_k = \sum_{i=1}^k \frac{1}{i(i+1)}
   \]
   This sum can be simplified using partial fractions:
   \[
   \frac{1}{i(i+1)} = \frac{1}{i} - \frac{1}{i+1}
   \]
   Therefore:
   \[
   \sum_{i=1}^k \left( \frac{1}{i} - \frac{1}{i+1} \right) = 1 - \frac{1}{k+1} = \frac{k}{k+1}
   \]
   Thus:
   \[
   a_1 + a_2 + \cdots + a_k = \frac{k}{k+1}
   \]

4. **Inductive Step:**
   We need to show that:
   \[
   a_1 + a_2 + \cdots + a_k + a_{k+1} = (k+1)^2 a_{k+1}
   \]
   Using the previous result:
   \[
   \frac{k}{k+1} + a_{k+1} = (k+1)^2 a_{k+1}
   \]
   Solving for \( a_{k+1} \):
   \[
   \frac{k}{k+1} + a_{k+1} = (k+1)^2 a_{k+1}
   \]
   \[
   \frac{k}{k+1} = (k+1)^2 a_{k+1} - a_{k+1}
   \]
   \[
   \frac{k}{k+1} = a_{k+1} \left( (k+1)^2 - 1 \right)
   \]
   \[
   \frac{k}{k+1} = a_{k+1} \left( k^2 + 2k + 1 - 1 \right)
   \]
   \[
   \frac{k}{k+1} = a_{k+1} (k^2 + 2k)
   \]
   \[
   a_{k+1} = \frac{k}{k+1} \cdot \frac{1}{k^2 + 2k}
   \]
   \[
   a_{k+1} = \frac{k}{k+1} \cdot \frac{1}{k(k+2)}
   \]
   \[
   a_{k+1} = \frac{1}{(k+1)(k+2)}
   \]
   This completes the inductive step.

5. **Conclusion:**
   By mathematical induction, we have shown that for all \( n \geq 1 \):
   \[
   a_n = \frac{1}{n(n+1)}
   \]

The final answer is \( \boxed{ a_n = \frac{1}{n(n+1)} } \).