For real numbers $B,M,$ and $T,$ we have $B^2+M^2+T^2 =2022$ and $B+M+T =72.$ Compute the sum of the minimum and maximum possible values of $T.$
1. We start with the given equations:
   \[
   B^2 + M^2 + T^2 = 2022
   \]
   \[
   B + M + T = 72
   \]
   We need to find the sum of the minimum and maximum possible values of \( T \).

2. Express \( B + M \) and \( B^2 + M^2 \) in terms of \( T \):
   \[
   B + M = 72 - T
   \]
   \[
   B^2 + M^2 = 2022 - T^2
   \]

3. Use the identity for the square of a sum:
   \[
   (B + M)^2 = B^2 + M^2 + 2BM
   \]
   Substituting the known values:
   \[
   (72 - T)^2 = 2022 - T^2 + 2BM
   \]

4. Rearrange to solve for \( 2BM \):
   \[
   (72 - T)^2 = 2022 - T^2 + 2BM
   \]
   \[
   (72 - T)^2 - 2022 + T^2 = 2BM
   \]
   \[
   5184 - 144T + T^2 - 2022 + T^2 = 2BM
   \]
   \[
   2T^2 - 144T + 3162 = 2BM
   \]
   \[
   T^2 - 72T + 1581 = BM
   \]

5. Apply the AM-GM inequality to \( B \) and \( M \):
   \[
   \frac{B + M}{2} \geq \sqrt{BM}
   \]
   \[
   \frac{72 - T}{2} \geq \sqrt{BM}
   \]
   \[
   \left( \frac{72 - T}{2} \right)^2 \geq BM
   \]
   \[
   \frac{(72 - T)^2}{4} \geq BM
   \]

6. Substitute \( BM \) from the previous step:
   \[
   T^2 - 72T + 1581 \leq \frac{(72 - T)^2}{4}
   \]

7. Simplify the inequality:
   \[
   4(T^2 - 72T + 1581) \leq (72 - T)^2
   \]
   \[
   4T^2 - 288T + 6324 \leq 5184 - 144T + T^2
   \]
   \[
   4T^2 - 288T + 6324 \leq T^2 - 144T + 5184
   \]
   \[
   3T^2 - 144T + 1140 \leq 0
   \]

8. Solve the quadratic inequality:
   \[
   3T^2 - 144T + 1140 \leq 0
   \]
   \[
   T^2 - 48T + 380 \leq 0
   \]

9. Factorize the quadratic expression:
   \[
   (T - 10)(T - 38) \leq 0
   \]

10. Determine the range of \( T \):
    \[
    10 \leq T \leq 38
    \]

11. The minimum value of \( T \) is 10 and the maximum value is 38. The sum of these values is:
    \[
    10 + 38 = 48
    \]

The final answer is \(\boxed{48}\)