Let $A_n$ denote the answer to the $n$th problem on this contest ($n=1,\dots,30$); in particular, the answer to this problem is $A_1$. Compute $2A_1(A_1+A_2+\dots+A_{30})$.

[i]Proposed by Yang Liu[/i]
1. We are given that \( A_1 \) is the answer to the first problem, and we need to compute \( 2A_1(A_1 + A_2 + \dots + A_{30}) \).

2. Let's denote the sum of all answers from \( A_1 \) to \( A_{30} \) as \( S \). Therefore, we can write:
   \[
   S = A_1 + A_2 + \dots + A_{30}
   \]

3. The expression we need to compute is:
   \[
   2A_1(A_1 + A_2 + \dots + A_{30}) = 2A_1 S
   \]

4. Given that \( A_1 \) is the answer to the first problem, we substitute \( A_1 \) into the expression:
   \[
   2A_1(A_1 + A_2 + \dots + A_{30}) = 2A_1 S
   \]

5. We need to consider the possible values of \( A_1 \). Since \( A_1 \) is a nonnegative integer (as all OMO answers are nonnegative integers), we have two cases:
   - Case 1: \( A_1 = 0 \)
   - Case 2: \( A_1 \neq 0 \)

6. **Case 1: \( A_1 = 0 \)**
   \[
   2A_1 S = 2 \cdot 0 \cdot S = 0
   \]
   In this case, the expression evaluates to 0.

7. **Case 2: \( A_1 \neq 0 \)**
   \[
   2A_1 S = 2A_1(A_1 + A_2 + \dots + A_{30})
   \]
   Since \( A_1 \neq 0 \), we can divide both sides by \( 2A_1 \):
   \[
   1 = A_1 + A_2 + \dots + A_{30}
   \]
   This implies that \( S = \frac{1}{2} \), which is not possible because \( S \) must be an integer (as all \( A_i \) are nonnegative integers).

8. Therefore, the only valid solution is when \( A_1 = 0 \).

Conclusion:
\[
\boxed{0}
\]