Let $t$ be TNYWR.

The sum of two numbers is $t$ and the positive difference between the squares of these two numbers is 208. What is the larger of the two numbers?
Let the numbers be $a$ and $b$.

We have that $a+b=t$ and $a^{2}-b^{2}=208$.

Factoring the difference of squares, we have $(a-b)(a+b)=208$, so $t(a-b)=208$, which can be solved for $a-b$ to get $a-b=\frac{208}{t}$.

Adding this equation to $a+b=t$, we get $2 a=t+\frac{208}{t}$ so $a=\frac{t}{2}+\frac{104}{t}$.

Substituting $t=104$, we get $a=\frac{104}{2}+\frac{104}{104}=53$. Observe that since $(a-b)(a+b)=208$ and $a+b$ and 208 are positive, we have that $a-b>0$ so $a>b$.

Therefore, $a$ is the largest of the two numbers so the answer is 53 .

ANSWER: $(52,104,53)$