An interesting two digit number

Two-digit squares: $16, 25, 36, 49, 64, 81$ .

Since $25, 36$ don't have a square number, we can eliminate them.

Now we have left: $16, 49, 64, 81$ .

Since $16, 64, 81$ have only one square number, we can eliminate them.

Therefore, there is only $\fbox 1$ number that satisfies @Alak Bhattacharya 's conditions: $\fbox {49}$

Therefore, the answer is $\fbox {49 + 1 = 50}$

All two-digit perfect squares are:

$16, 25, 36, 49, 64, 81$

From these squares either the first or second digit is not a perfect square:

$\cancel{16}, \cancel{25}, \cancel{36}, {\color{#3D99F6}{49}}, \cancel{64}, \cancel{81}$

The answer is $=$ Number of such numbers + Sum of all such Numbers:

$49 + 1 = \boxed{50}$

If the digits are perfect squares, the tens digit must be $1$ , $4$ , or $9$ .

The 2-digit perfect square with the leading digit $1$ is $16$ , which does not satisfy.

The 2-digit perfect square with the leading digit $4$ is $49$ , which does satisfy.

There are no 2-digit perfect squares with the leading digit $9$ .

Therefore, there is $1$ perfect square that satisfies the conditions, and it is $49$ . The sum is therefore $1 + 49 = \boxed{50}$ .