Find the number of ways in which three numbers can be selected from the set $\{1,2,\cdots ,4n\}$, such that the sum of the three selected numbers is divisible by $4$.
1. **Break the set into parts based on congruency modulo 4:**
   The set $\{1, 2, \cdots, 4n\}$ can be divided into four subsets:
   \[
   S_1 = \{1, 5, \cdots, 4n-3\}, \quad S_2 = \{2, 6, \cdots, 4n-2\}, \quad S_3 = \{3, 7, \cdots, 4n-1\}, \quad S_4 = \{4, 8, \cdots, 4n\}
   \]
   Each element of $S_i$ is congruent to $i \pmod{4}$. The cardinality of each set is $n$.

2. **Identify the combinations of sets that sum to a multiple of 4:**
   We need to find combinations of three numbers such that their sum is divisible by 4. The possible combinations are:
   - $a_4 + a_4 + a_4 \equiv 0 \pmod{4}$
   - $a_2 + a_2 + a_4 \equiv 0 \pmod{4}$
   - $a_1 + a_1 + a_2 \equiv 0 \pmod{4}$
   - $a_3 + a_1 + a_4 \equiv 0 \pmod{4}$
   - $a_3 + a_3 + a_2 \equiv 0 \pmod{4}$

3. **Calculate the number of ways to choose the numbers for each combination:**
   - For $a_4 + a_4 + a_4$: 
     \[
     \binom{n}{3}
     \]
   - For $a_2 + a_2 + a_4$: 
     \[
     \binom{n}{2} \cdot \binom{n}{1}
     \]
   - For $a_1 + a_1 + a_2$: 
     \[
     \binom{n}{2} \cdot \binom{n}{1}
     \]
   - For $a_3 + a_1 + a_4$: 
     \[
     \binom{n}{1} \cdot \binom{n}{1} \cdot \binom{n}{1}
     \]
   - For $a_3 + a_3 + a_2$: 
     \[
     \binom{n}{2} \cdot \binom{n}{1}
     \]

4. **Sum the number of ways for all combinations:**
   \[
   \binom{n}{3} + 3 \left( \binom{n}{2} \cdot \binom{n}{1} \right) + \left( \binom{n}{1} \cdot \binom{n}{1} \cdot \binom{n}{1} \right)
   \]
   Simplifying the binomial coefficients:
   \[
   \binom{n}{3} = \frac{n(n-1)(n-2)}{6}, \quad \binom{n}{2} = \frac{n(n-1)}{2}, \quad \binom{n}{1} = n
   \]
   Therefore:
   \[
   \binom{n}{3} + 3 \left( \frac{n(n-1)}{2} \cdot n \right) + n^3
   \]
   Simplifying further:
   \[
   \frac{n(n-1)(n-2)}{6} + 3 \left( \frac{n^2(n-1)}{2} \right) + n^3
   \]
   \[
   = \frac{n(n-1)(n-2)}{6} + \frac{3n^3(n-1)}{2} + n^3
   \]
   \[
   = \frac{n(n-1)(n-2)}{6} + \frac{3n^3(n-1)}{2} + n^3
   \]
   \[
   = \frac{n(n-1)(n-2) + 9n^3(n-1) + 6n^3}{6}
   \]
   \[
   = \frac{n(n-1)(n-2) + 9n^4 - 9n^3 + 6n^3}{6}
   \]
   \[
   = \frac{n(n-1)(n-2) + 9n^4 - 3n^3}{6}
   \]
   \[
   = \frac{n^3 - 3n^2 + 2n + 9n^4 - 3n^3}{6}
   \]
   \[
   = \frac{9n^4 - 2n^3 - 3n^2 + 2n}{6}
   \]

5. **Verify the final expression:**
   The final expression should match $\frac{1}{4} \binom{4n}{3}$:
   \[
   \binom{4n}{3} = \frac{4n(4n-1)(4n-2)}{6}
   \]
   \[
   = \frac{4n \cdot 4n \cdot 4n}{6} = \frac{64n^3}{6} = \frac{32n^3}{3}
   \]
   \[
   \frac{1}{4} \binom{4n}{3} = \frac{1}{4} \cdot \frac{32n^3}{3} = \frac{8n^3}{3}
   \]

The final answer is $\boxed{n^3 + 3n \binom{n}{2} + \binom{n}{3}}$.