To solve the problem, we start by finding the proper divisors of $284$.

Step 1: Find the divisors of $284$

We first find the prime factorization of $284$:

$284 = 2^2 \times 71$

Step 2: List the divisors of $284$

The divisors can be calculated using the formula for the number of divisors based on prime factorization. The divisors of $284$ are:

1. $1$ (which is $2^0 \times 71^0$)
2. $2$ (which is $2^1 \times 71^0$)
3. $4$ (which is $2^2 \times 71^0$)
4. $71$ (which is $2^0 \times 71^1$)
5. $142$ (which is $2^1 \times 71^1$)
6. $284$ (which is $2^2 \times 71^1$)

Thus, the complete set of divisors is $\{1, 2, 4, 71, 142, 284\}$.

Step 3: Identify the proper divisors of $284$

The proper divisors are the divisors excluding $284$ itself. Therefore, the proper divisors are:

$1, 2, 4, 71, 142$

Step 4: Calculate the sum of the proper divisors of $284$

Now, we compute the sum of these proper divisors:

$1 + 2 + 4 + 71 + 142 = 220$

Step 5: Find the proper divisors of $220$

Next, we need to find the proper divisors of $220$. First, we factor $220$:

$220 = 2^2 \times 5 \times 11$

Step 6: List the divisors of $220$

The divisors of $220$ can be found as follows:

1. $1$ (which is $2^0 \times 5^0 \times 11^0$)
2. $2$ (which is $2^1 \times 5^0 \times 11^0$)
3. $4$ (which is $2^2 \times 5^0 \times 11^0$)
4. $5$ (which is $2^0 \times 5^1 \times 11^0$)
5. $10$ (which is $2^1 \times 5^1 \times 11^0$)
6. $20$ (which is $2^2 \times 5^1 \times 11^0$)
7. $11$ (which is $2^0 \times 5^0 \times 11^1$)
8. $22$ (which is $2^1 \times 5^0 \times 11^1$)
9. $44$ (which is $2^2 \times 5^0 \times 11^1$)
10. $55$ (which is $2^0 \times 5^1 \times 11^1$)
11. $110$ (which is $2^1 \times 5^1 \times 11^1$)
12. $220$ (which is $2^2 \times 5^1 \times 11^1$)

Thus, the complete set of divisors is $\{1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110, 220\}$.

Step 7: Identify the proper divisors of $220$

The proper divisors of $220$ (excluding $220$) are:

$1, 2, 4, 5, 10, 11, 20, 22, 44, 55, 110$

Step 8: Calculate the sum of the proper divisors of $220$

Now, we compute the sum of these proper divisors:

$1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110 = 284$

Conclusion

Thus, the sum of the proper divisors of the sum of the proper divisors of $284$ is

$\boxed{284}$