We say that an integer $x\in\{1,\dots,102\}$ is $\textit{square-ish}$ if there exists some integer $n$ such that $x\equiv n^2+n\pmod{103}$. Compute the product of all $\textit{square-ish}$ integers modulo $103$.
1. **Understanding the Problem:**
   We need to find the product of all integers \( x \in \{1, \dots, 102\} \) that can be expressed as \( x \equiv n^2 + n \pmod{103} \) for some integer \( n \).

2. **Analyzing the Condition:**
   The condition \( x \equiv n^2 + n \pmod{103} \) can be rewritten as:
   \[
   x \equiv n(n + 1) \pmod{103}
   \]
   We need to determine the product of all such \( x \).

3. **Identifying Distinct Values:**
   For distinct \( m \) and \( n \), the values \( m^2 + m \) and \( n^2 + n \) are distinct modulo 103 if \( m + n \not\equiv -1 \pmod{103} \). This means that each \( n \) from 0 to 102 will give a unique value of \( n(n + 1) \) modulo 103.

4. **Range of \( n \):**
   Since \( n \) ranges from 0 to 102, we need to consider the product of \( n(n + 1) \) for \( n = 0, 1, 2, \ldots, 102 \).

5. **Simplifying the Product:**
   The product of all \( n(n + 1) \) for \( n = 0 \) to \( 102 \) can be written as:
   \[
   \prod_{n=0}^{102} n(n + 1)
   \]
   This can be split into two separate products:
   \[
   \left( \prod_{n=0}^{102} n \right) \left( \prod_{n=0}^{102} (n + 1) \right)
   \]

6. **Calculating the Factorials:**
   The first product is the factorial of 102:
   \[
   \prod_{n=0}^{102} n = 102!
   \]
   The second product is the factorial of 103:
   \[
   \prod_{n=0}^{102} (n + 1) = 103!
   \]

7. **Using Wilson's Theorem:**
   Wilson's Theorem states that for a prime \( p \):
   \[
   (p-1)! \equiv -1 \pmod{p}
   \]
   Applying this to 103, we get:
   \[
   102! \equiv -1 \pmod{103}
   \]
   And since \( 103! = 103 \cdot 102! \):
   \[
   103! \equiv 0 \cdot (-1) \equiv 0 \pmod{103}
   \]

8. **Combining the Results:**
   The product of all \( n(n + 1) \) modulo 103 is:
   \[
   102! \cdot 103! \equiv (-1) \cdot 0 \equiv 0 \pmod{103}
   \]

The final answer is \(\boxed{0}\).