Answer: 80

Small Hint:

The condition $a + d \gt b + c$ is the same as $d - c \gt b - a$

Big Hint:

The map $(a, b, c, d)$ $\small \mapsto (11 - d,\, 11 - c,\, 11 - b,\, 11 - a)$ swaps $d - c$ with $b - a;$ count the quadruples with $d - c = b - a$ and use symmetry

Solution:

The condition $a + d \gt b + c$ is equivalent to $d - c \gt b - a.$ There are $\binom{10}{4} = 210$ quadruples in all, and the involution $(a, b, c, d)$ $\small \mapsto (11 - d,\, 11 - c,\, 11 - b,\, 11 - a)$ exchanges the outer gaps $b - a$ and $d - c.$ So the quadruples with $d - c \gt b - a$ and those with $d - c \lt b - a$ are equinumerous, and the answer is $\frac{210 - T}{2},$ where $T$ counts quadruples with $d - c = b - a.$

If $b - a = d - c = k$ and $c - b = j,$ the quadruple is determined by $(a, j, k)$ with $a, j, k \ge 1$ and $a + 2k + j \le 10.$ For $k = 1,$ $2,$ $3,$ and $4,$ respectively, the bounds on $a + j$ are $8,$ $6,$ $4,$ and $2;$ the corresponding counts are $28,$ $15,$ $6,$ and $1,$ so $T = 50.$

Therefore the number of interesting quadruples is $\frac{210 - 50}{2} = 80.$