What is the smallest positive integer $n$ such that there exists a choice of signs for which \[1^2\pm2^2\pm3^2\ldots\pm n^2=0\] is true?

[i]2019 CCA Math Bonanza Team Round #5[/i]
To find the smallest positive integer \( n \) such that there exists a choice of signs for which 
\[ 1^2 \pm 2^2 \pm 3^2 \pm \ldots \pm n^2 = 0 \]
is true, we need to consider the sum of the squares of the first \( n \) integers.

1. **Sum of Squares Formula**:
   The sum of the squares of the first \( n \) integers is given by:
   \[
   S = \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}
   \]

2. **Condition for Equal Partition**:
   We need to partition the sum \( S \) into two equal parts with appropriate signs. Therefore, \( S \) must be even:
   \[
   \frac{n(n+1)(2n+1)}{6} \text{ must be even}
   \]

3. **Checking Small Values of \( n \)**:
   We will check the values of \( n \) starting from 1 to find the smallest \( n \) that satisfies the condition.

   - For \( n = 1 \):
     \[
     S = \frac{1(1+1)(2 \cdot 1 + 1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1 \quad (\text{odd})
     \]
     Not possible.

   - For \( n = 2 \):
     \[
     S = \frac{2(2+1)(2 \cdot 2 + 1)}{6} = \frac{2 \cdot 3 \cdot 5}{6} = 5 \quad (\text{odd})
     \]
     Not possible.

   - For \( n = 3 \):
     \[
     S = \frac{3(3+1)(2 \cdot 3 + 1)}{6} = \frac{3 \cdot 4 \cdot 7}{6} = 14 \quad (\text{even})
     \]
     Possible, but we need to check if we can partition it.

   - For \( n = 4 \):
     \[
     S = \frac{4(4+1)(2 \cdot 4 + 1)}{6} = \frac{4 \cdot 5 \cdot 9}{6} = 30 \quad (\text{even})
     \]
     Possible, but we need to check if we can partition it.

   - For \( n = 5 \):
     \[
     S = \frac{5(5+1)(2 \cdot 5 + 1)}{6} = \frac{5 \cdot 6 \cdot 11}{6} = 55 \quad (\text{odd})
     \]
     Not possible.

   - For \( n = 6 \):
     \[
     S = \frac{6(6+1)(2 \cdot 6 + 1)}{6} = \frac{6 \cdot 7 \cdot 13}{6} = 91 \quad (\text{odd})
     \]
     Not possible.

   - For \( n = 7 \):
     \[
     S = \frac{7(7+1)(2 \cdot 7 + 1)}{6} = \frac{7 \cdot 8 \cdot 15}{6} = 140 \quad (\text{even})
     \]
     Possible, and we need to check if we can partition it.

4. **Verification for \( n = 7 \)**:
   We need to check if we can partition the sum \( S = 140 \) into two equal parts:
   \[
   1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 = 1 + 4 + 9 + 16 + 25 + 36 + 49 = 140
   \]
   We need to find a subset of these squares that sums to \( 70 \):
   \[
   1 + 4 + 9 + 16 + 36 + 4 = 70
   \]
   This is possible, so \( n = 7 \) works.

The final answer is \( \boxed{7} \).