Just 10 ants

$v_k=\frac {(11-k)^2-k^2}{1}$ » velocity of the $k^{th}$ ant.

Thus, distance of the $k^{th}$ ant (from 0) $x_k=x_0 + v_k \cdot t, x_0= k^2$

At a certain time, let the $k^{th}$ and $j^{th}$ ants meet.

$x_k=x_j$

$k^2 + 121 - 22kt= j^2 + 121 - 22jt$

$t=\frac {k+j}{22}$

Putting k = 1, we get $t=\frac {3}{22},\frac {4}{22}, \frac {5}{22}...\frac {11}{22}$ (9 values), for all the other j's. ( j = 2, 3, 4...., 10 ).

Putting k = 2, we get $t = \frac {5}{22}, \frac {6}{22},....\frac {12}{22}$ , (j=3,4,5....,10)

Thus, we observe that each new $k$ will give us just one more new distinct value of $t$ .

We got 9 values from k=1, and one more from each k=2,3,4....,9 (8 more values). Hence the answer is $9+8=$ $17$