Suppose $w,x,y,z$ satisfy \begin{align*}w+x+y+z&=25,\\wx+wy+wz+xy+xz+yz&=2y+2z+193\end{align*} The largest possible value of $w$ can be expressed in lowest terms as $w_1/w_2$ for some integers $w_1,w_2>0$.  Find $w_1+w_2$.
1. We start with the given equations:
   \[
   w + x + y + z = 25
   \]
   \[
   wx + wy + wz + xy + xz + yz = 2y + 2z + 193
   \]

2. We introduce new variables to simplify the problem. Let:
   \[
   a = y + 2 \quad \text{and} \quad b = z + 2
   \]
   Then:
   \[
   w + x + a + b = 29
   \]

3. We square the sum of the variables:
   \[
   (w + x + a + b)^2 = 29^2 = 841
   \]
   Expanding the square, we get:
   \[
   w^2 + x^2 + a^2 + b^2 + 2(wx + wa + wb + xa + xb + ab) = 841
   \]

4. We know from the problem statement that:
   \[
   wx + wy + wz + xy + xz + yz = 2y + 2z + 193
   \]
   Substituting \(a = y + 2\) and \(b = z + 2\), we get:
   \[
   wx + w(y+2) + w(z+2) + x(y+2) + x(z+2) + (y+2)(z+2) = 2(y+2) + 2(z+2) + 193
   \]
   Simplifying, we have:
   \[
   wx + wy + wz + 2w + xy + xz + 2x + yz + 2y + 2z + 4 = 2y + 2z + 201
   \]
   \[
   wx + wy + wz + xy + xz + yz + 2w + 2x + 4 = 201
   \]

5. We use the Cauchy-Schwarz inequality to find the maximum value of \(w\):
   \[
   (w^2 + x^2 + a^2 + b^2)(1^2 + 1^2 + 1^2 + 1^2) \geq (w + x + a + b)^2
   \]
   \[
   (w^2 + x^2 + (y+2)^2 + (z+2)^2) \cdot 4 \geq 29^2
   \]
   \[
   4(w^2 + x^2 + y^2 + 4y + 4 + z^2 + 4z + 4) \geq 841
   \]
   \[
   4(w^2 + x^2 + y^2 + z^2 + 4y + 4z + 8) \geq 841
   \]
   \[
   4(w^2 + x^2 + y^2 + z^2 + 4y + 4z + 8) = 4(247 - w^2) \geq 841
   \]
   \[
   988 - 4w^2 \geq 841
   \]
   \[
   147 \geq 4w^2
   \]
   \[
   36.75 \geq w^2
   \]
   \[
   w \leq \frac{25}{2}
   \]

6. We solve the quadratic inequality:
   \[
   0 \geq 4w^2 - 58w + 100
   \]
   \[
   0 \geq (2w - 25)(w - 2)
   \]
   \[
   2 \leq w \leq \frac{25}{2}
   \]

7. The largest possible value of \(w\) is \(\frac{25}{2}\). Therefore, \(w_1 = 25\) and \(w_2 = 2\).

The final answer is \( \boxed{ 25 + 2 = 27 } \).