Let $A,B,C,...,Z$ be $26$ nonzero real numbers. Suppose that $T=TNYWR$. Compute the smallest possible value of \[
	  \left\lceil A^2+B^2+\cdots+Z^2\right\rceil .
\] (The notation $\left\lceil x\right\rceil$ denotes the least integer $n$ such that $n\geq x$.)

[i]Proposed by Luke Robitaille[/i]
To find the smallest possible value of \(\left\lceil A^2 + B^2 + \cdots + Z^2 \right\rceil\), we need to minimize the sum of the squares of 26 nonzero real numbers. 

1. **Using the Arithmetic Mean-Geometric Mean Inequality (AM-GM):**
   The AM-GM inequality states that for any non-negative real numbers \(x_1, x_2, \ldots, x_n\),
   \[
   \frac{x_1 + x_2 + \cdots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \cdots x_n}
   \]
   with equality if and only if \(x_1 = x_2 = \cdots = x_n\).

2. **Applying AM-GM to our problem:**
   Let \(A, B, C, \ldots, Z\) be the 26 nonzero real numbers. We want to minimize \(A^2 + B^2 + \cdots + Z^2\). By AM-GM,
   \[
   \frac{A^2 + B^2 + \cdots + Z^2}{26} \geq \sqrt[26]{A^2 B^2 \cdots Z^2}
   \]
   Since \(A, B, C, \ldots, Z\) are nonzero real numbers, we can assume without loss of generality that their product is 1 (i.e., \(A \cdot B \cdot C \cdots \cdot Z = 1\)). Therefore,
   \[
   \sqrt[26]{A^2 B^2 \cdots Z^2} = \sqrt[26]{(A B \cdots Z)^2} = \sqrt[26]{1^2} = 1
   \]
   Thus,
   \[
   \frac{A^2 + B^2 + \cdots + Z^2}{26} \geq 1 \implies A^2 + B^2 + \cdots + Z^2 \geq 26
   \]

3. **Checking if the minimum value can be achieved:**
   Equality in AM-GM holds if and only if \(A^2 = B^2 = \cdots = Z^2\). Let \(A = B = \cdots = Z = 1\). Then,
   \[
   A^2 + B^2 + \cdots + Z^2 = 1^2 + 1^2 + \cdots + 1^2 = 26
   \]

4. **Ceiling function:**
   The ceiling function \(\left\lceil x \right\rceil\) gives the smallest integer greater than or equal to \(x\). Since \(A^2 + B^2 + \cdots + Z^2 = 26\) is already an integer,
   \[
   \left\lceil 26 \right\rceil = 26
   \]

The final answer is \(\boxed{26}\).