There exists a unique strictly increasing arithmetic sequence $\{a_i\}_{i=1}^{100}$ of positive integers such that \[a_1+a_4+a_9+\cdots+a_{100}=\text{1000},\] where the summation runs over all terms of the form $a_{i^2}$ for $1\leq i\leq 10$.  Find $a_{50}$.

[i]Proposed by David Altizio and Tony Kim[/i]
1. Let \( a_1 = a \) and \( a_{k+1} - a_k = d \). This means the sequence is an arithmetic sequence with the first term \( a \) and common difference \( d \).

2. We are given the condition:
   \[
   a_1 + a_4 + a_9 + \cdots + a_{100} = 1000
   \]
   where the summation runs over all terms of the form \( a_{i^2} \) for \( 1 \leq i \leq 10 \).

3. Each term \( a_{i^2} \) in the sequence can be written as:
   \[
   a_{i^2} = a + (i^2 - 1)d
   \]

4. Therefore, the sum of these terms is:
   \[
   \sum_{i=1}^{10} a_{i^2} = \sum_{i=1}^{10} \left( a + (i^2 - 1)d \right)
   \]

5. Simplifying the sum, we get:
   \[
   \sum_{i=1}^{10} a_{i^2} = \sum_{i=1}^{10} a + \sum_{i=1}^{10} (i^2 - 1)d
   \]
   \[
   = 10a + d \sum_{i=1}^{10} (i^2 - 1)
   \]

6. Calculate the sum of \( i^2 - 1 \) for \( i \) from 1 to 10:
   \[
   \sum_{i=1}^{10} (i^2 - 1) = \sum_{i=1}^{10} i^2 - \sum_{i=1}^{10} 1
   \]
   \[
   = \sum_{i=1}^{10} i^2 - 10
   \]

7. The sum of the squares of the first 10 natural numbers is:
   \[
   \sum_{i=1}^{10} i^2 = 1^2 + 2^2 + 3^2 + \cdots + 10^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385
   \]

8. Therefore:
   \[
   \sum_{i=1}^{10} (i^2 - 1) = 385 - 10 = 375
   \]

9. Substituting back into the sum equation:
   \[
   10a + 375d = 1000
   \]

10. Simplify the equation:
    \[
    2a + 75d = 200
    \]

11. Solving for \( a \) and \( d \), we can try \( d = 2 \):
    \[
    2a + 75 \cdot 2 = 200
    \]
    \[
    2a + 150 = 200
    \]
    \[
    2a = 50
    \]
    \[
    a = 25
    \]

12. Now, we need to find \( a_{50} \):
    \[
    a_{50} = a + 49d
    \]
    \[
    = 25 + 49 \cdot 2
    \]
    \[
    = 25 + 98
    \]
    \[
    = 123
    \]

The final answer is \( \boxed{123} \).