The sequence $(a_n)$ is determined by $a_1 = 0$ and
$(n+1)^3a_{n+1} = 2n^2(2n+1)a_n+2(3n+1)$ for $n \geq 1$.
Prove that infinitely many terms of the sequence are positive integers.
1. **Define a new sequence \( b_n \):**
   Let \( b_n = n^2 a_n \). The given recurrence relation is:
   \[
   (n+1)^3 a_{n+1} = 2n^2 (2n+1) a_n + 2(3n+1)
   \]
   Substituting \( a_n = \frac{b_n}{n^2} \) and \( a_{n+1} = \frac{b_{n+1}}{(n+1)^2} \), we get:
   \[
   (n+1)^3 \frac{b_{n+1}}{(n+1)^2} = 2n^2 (2n+1) \frac{b_n}{n^2} + 2(3n+1)
   \]
   Simplifying, we obtain:
   \[
   (n+1) b_{n+1} = 2(2n+1) b_n + 2(3n+1)
   \]

2. **Define another new sequence \( c_n \):**
   Let \( c_n = b_n + 2 \). Then:
   \[
   (n+1) c_{n+1} = (n+1) (b_{n+1} + 2) = (n+1) b_{n+1} + 2(n+1)
   \]
   Using the previous result:
   \[
   (n+1) b_{n+1} = 2(2n+1) b_n + 2(3n+1)
   \]
   We get:
   \[
   (n+1) c_{n+1} = 2(2n+1) b_n + 2(3n+1) + 2(n+1)
   \]
   Simplifying:
   \[
   (n+1) c_{n+1} = 2(2n+1) b_n + 2(3n+1) + 2(n+1) = 2(2n+1) (b_n + 2) = 2(2n+1) c_n
   \]
   Therefore, we have:
   \[
   (n+1) c_{n+1} = 2(2n+1) c_n
   \]

3. **Prove by induction that \( c_n = \binom{2n}{n} \):**
   - **Base case:** For \( n = 1 \):
     \[
     c_1 = b_1 + 2 = 0 + 2 = 2
     \]
     And:
     \[
     \binom{2 \cdot 1}{1} = \binom{2}{1} = 2
     \]
     So, \( c_1 = \binom{2}{1} \).

   - **Inductive step:** Assume \( c_k = \binom{2k}{k} \) for some \( k \geq 1 \). We need to show that \( c_{k+1} = \binom{2(k+1)}{k+1} \):
     \[
     (k+1) c_{k+1} = 2(2k+1) c_k
     \]
     By the inductive hypothesis:
     \[
     (k+1) c_{k+1} = 2(2k+1) \binom{2k}{k}
     \]
     Using the identity:
     \[
     \binom{2(k+1)}{k+1} = \frac{(2(k+1))!}{(k+1)! (k+1)!}
     \]
     We need to show:
     \[
     c_{k+1} = \frac{2(2k+1)}{k+1} \binom{2k}{k}
     \]
     Simplifying:
     \[
     \binom{2(k+1)}{k+1} = \frac{(2k+2)(2k+1)}{(k+1)(k+1)} \binom{2k}{k} = \frac{2(2k+1)}{k+1} \binom{2k}{k}
     \]
     Therefore, \( c_{k+1} = \binom{2(k+1)}{k+1} \).

4. **Express \( a_n \) in terms of \( c_n \):**
   Since \( c_n = b_n + 2 \) and \( b_n = n^2 a_n \), we have:
   \[
   c_n = n^2 a_n + 2
   \]
   Thus:
   \[
   n^2 a_n = c_n - 2 = \binom{2n}{n} - 2
   \]
   Therefore:
   \[
   a_n = \frac{1}{n^2} \left( \binom{2n}{n} - 2 \right)
   \]

5. **Show that \( \binom{2p}{p} - 2 \) is divisible by \( p^2 \) for prime \( p \):**
   For \( n = p \) where \( p \) is a prime, we use the identity:
   \[
   \binom{2p}{p} - 2 = \sum_{k=1}^{p-1} \binom{p}{k}^2
   \]
   Since each term \( \binom{p}{k} \) is an integer, the sum is an integer. Moreover, it is known that \( \binom{2p}{p} \equiv 2 \pmod{p^2} \), implying that \( \binom{2p}{p} - 2 \) is divisible by \( p^2 \).

   Therefore, \( a_p = \frac{1}{p^2} (\binom{2p}{p} - 2) \) is an integer for prime \( p \).

Conclusion:
\[
\boxed{\text{The sequence } (a_n) \text{ has infinitely many positive integer terms.}}
\]